Quiz: Projectile Motion Fundamentals — 62 questions

1. What makes projectile motion a two-dimensional type of motion?

A. The particle moves repeatedly between two fixed points on one axis

B. The particle moves simultaneously along independent horizontal and vertical axes

C. The particle remains stationary in one coordinate direction

D. The particle follows a straight path with changing speed

Answer: (B)

Explanation

Projectile motion is two-dimensional because its motion has simultaneous horizontal and vertical components. Straight-line motion is restricted to one axis, so it does not have this two-axis structure.

2. A projectile is being analyzed to determine its horizontal displacement; which motion can be treated independently for that calculation?

A. The horizontal motion can be ignored while analyzing the horizontal motion

B. The vertical motion can be ignored while analyzing the horizontal motion

C. The vertical component must be treated as the source of horizontal acceleration

D. Both components must be combined before either displacement can be found

Answer: (B)

Explanation

The independent-axis method allows horizontal motion to be analyzed without considering the vertical motion for that calculation. The vertical component is instead examined separately when finding vertical quantities.

3. In the standard projectile model, how is air resistance treated?

A. It is neglected unless the problem explicitly includes it

B. It is assigned a constant value for every projectile

C. It is included whenever the projectile has a horizontal component

D. It is treated as equal to the projectile's weight

Answer: (A)

Explanation

The standard model neglects air resistance unless the problem states that it must be considered. A problem that explicitly includes air resistance requires a different model.

4. What gravitational acceleration is normally used in the standard projectile model?

A. A constant value of $9.8\,\mathrm{m\,s^{-2}}$ or approximately $10\,\mathrm{m\,s^{-2}}$

B. A value that changes with the projectile's instantaneous speed

C. A value that decreases continuously as the projectile rises

D. A horizontal acceleration of $9.8\,\mathrm{m\,s^{-2}}$ during the flight

Answer: (A)

Explanation

The standard model treats gravity as constant, using $9.8\,\mathrm{m\,s^{-2}}$ or approximately $10\,\mathrm{m\,s^{-2}}$ unless another value is specified. Changes in gravitational acceleration are considered only when the problem specifically requires them.

5. A projectile is moving upward at an angle while air resistance is neglected; what is the direction of its acceleration?

A. Vertically upward with magnitude $g$

B. Vertically downward with magnitude $g$

C. Horizontally opposite to its horizontal velocity

D. Along the projectile's instantaneous direction of motion

Answer: (B)

Explanation

Under gravity alone, acceleration remains vertically downward with magnitude $g$ during rising, falling, or angled motion. The direction of velocity does not determine the direction of gravitational acceleration.

6. Why does a projectile's horizontal velocity remain constant before impact in the standard model?

A. Its horizontal acceleration is zero

B. Air resistance continually restores lost horizontal speed

C. Its vertical acceleration cancels its horizontal velocity

D. Gravity acts horizontally with constant magnitude

Answer: (A)

Explanation

With no horizontal acceleration, the horizontal component of velocity does not change throughout the flight. Gravity acts vertically, so it does not alter the horizontal velocity in this model.

7. A particle is projected at $100\,\mathrm{m\,s^{-1}}$ at $53^\circ$ to the vertical; what are its initial horizontal and vertical velocity components?

A. $u_x=80\,\mathrm{m\,s^{-1}}$ and $u_y=60\,\mathrm{m\,s^{-1}}$

B. $u_x=53\,\mathrm{m\,s^{-1}}$ and $u_y=100\,\mathrm{m\,s^{-1}}$

C. $u_x=100\,\mathrm{m\,s^{-1}}$ and $u_y=53\,\mathrm{m\,s^{-1}}$

D. $u_x=60\,\mathrm{m\,s^{-1}}$ and $u_y=80\,\mathrm{m\,s^{-1}}$

Answer: (A)

Explanation

The angle is $37^\circ$ to the horizontal, so the components are $u_x=100\cos37^\circ=80\,\mathrm{m\,s^{-1}}$ and $u_y=100\sin37^\circ=60\,\mathrm{m\,s^{-1}}$. Interchanging these values would incorrectly treat the given angle as measured from the horizontal.

8. For the example with initial components $u_x=80\,\mathrm{m\,s^{-1}}$ and $u_y=60\,\mathrm{m\,s^{-1}}$, what is the velocity after time $t$ when $g=10\,\mathrm{m\,s^{-2}}$?

A. $\vec v(t)=80\hat i+(60+10t)\hat j\,\mathrm{m\,s^{-1}}$

B. $\vec v(t)=(80-10t)\hat i+60\hat j\,\mathrm{m\,s^{-1}}$

C. $\vec v(t)=(80+10t)\hat i+(60-10t)\hat j\,\mathrm{m\,s^{-1}}$

D. $\vec v(t)=80\hat i+(60-10t)\hat j\,\mathrm{m\,s^{-1}}$

Answer: (D)

Explanation

The horizontal velocity stays at $80\,\mathrm{m\,s^{-1}}$, while the vertical velocity decreases by $10\,\mathrm{m\,s^{-1}}$ each second, giving the stated vector. The alternatives incorrectly change the horizontal component or make the vertical component increase.

9. For the same projectile launched from the origin, which position vector describes its location at time $t$?

A. $\vec r(t)=80t^2\hat i+(60t-5t)\hat j\,\mathrm{m}$

B. $\vec r(t)=80t\hat i+(60t-5t^2)\hat j\,\mathrm{m}$

C. $\vec r(t)=(80-5t^2)\hat i+60t\hat j\,\mathrm{m}$

D. $\vec r(t)=80t\hat i+(60t+5t^2)\hat j\,\mathrm{m}$

Answer: (B)

Explanation

The horizontal coordinate is $80t$ because horizontal velocity is constant, while the vertical coordinate is $60t-5t^2$ because gravity contributes the quadratic term. The other expressions place the gravitational term in the wrong component or use an incorrect time dependence.

10. For the example with $u_y=60\,\mathrm{m\,s^{-1}}$ and $g=10\,\mathrm{m\,s^{-2}}$, how long does the projectile take to reach its highest point?

A. $5\,\mathrm{s}$

B. $6\,\mathrm{s}$

C. $10\,\mathrm{s}$

D. $12\,\mathrm{s}$

Answer: (B)

Explanation

At the highest point, the vertical velocity is zero, so $t_{\mathrm{max}}=u_y/g=60/10=6\,\mathrm{s}$. The value $12\,\mathrm{s}$ is the total flight time for a return to the same level, not the time to the peak.

11. For a projectile launched with speed $u$ at angle $\theta$, which expression correctly gives its initial velocity and acceleration?

A. $\vec{u}=u\sin\theta\,\hat{i}+u\cos\theta\,\hat{j}$ and $\vec{a}=-g\hat{i}$

B. $\vec{u}=u\cos\theta\,\hat{i}+u\sin\theta\,\hat{j}$ and $\vec{a}=-g\hat{j}$

C. $\vec{u}=u\,\hat{i}+u\,\hat{j}$ and $\vec{a}=-g\hat{i}$

D. $\vec{u}=u\cos\theta\,\hat{i}-u\sin\theta\,\hat{j}$ and $\vec{a}=g\hat{j}$

Answer: (B)

Explanation

Resolving the launch velocity into horizontal and vertical components gives $u\cos\theta$ and $u\sin\theta$, while gravity provides the downward acceleration $-g\hat{j}$. The horizontal component is constant, whereas the vertical component changes during the motion.

12. A projectile is launched with speed $u$ at angle $\theta$. What is its velocity after time $t$?

A. $\vec{v}=u\cos\theta\,\hat{i}+(u\sin\theta-gt)\,\hat{j}$

B. $\vec{v}=(u\cos\theta+gt)\,\hat{i}-u\sin\theta\,\hat{j}$

C. $\vec{v}=(u\cos\theta-gt)\,\hat{i}+u\sin\theta\,\hat{j}$

D. $\vec{v}=u\cos\theta\,\hat{i}+(u\sin\theta+gt)\,\hat{j}$

Answer: (A)

Explanation

Gravity changes the vertical velocity by $-gt$ while leaving the horizontal velocity at $u\cos\theta$. The expression with gravity affecting the horizontal component applies the acceleration in the wrong direction.

13. Which equation describes the trajectory of a projectile after eliminating time from its horizontal and vertical coordinates?

A. $y=x\cot\theta-\frac{gx^2}{2u^2\sin^2\theta}$

B. $y=xtan\theta-\frac{g x^2}{2u\cos\theta}$

C. $y=x\tan\theta+\frac{gx^2}{2u^2\cos^2\theta}$

D. $y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}$

Answer: (D)

Explanation

Eliminating time produces a linear launch term and a downward quadratic term, yielding the stated parabolic trajectory equation. A positive quadratic term would represent upward curvature rather than the effect of gravity.

14. For a projectile returning to its launch level, which pair gives its maximum height and horizontal range?

A. $H=\frac{u^2\sin\theta}{2g}$ and $R=\frac{u^2\sin^2\theta}{g}$

B. $H=\frac{u^2\sin^2\theta}{2g}$ and $R=\frac{u^2\sin 2\theta}{g}$

C. $H=\frac{u^2\cos^2\theta}{2g}$ and $R=\frac{2u^2\sin\theta}{g}$

D. $H=\frac{u\sin^2\theta}{2g}$ and $R=\frac{u^2\cos 2\theta}{g}$

Answer: (B)

Explanation

The vertical launch component determines the maximum height, giving $H=\frac{u^2\sin^2\theta}{2g}$, while the range is $R=\frac{u^2\sin 2\theta}{g}$. The alternatives use incorrect trigonometric factors or incorrect powers of the launch speed.

15. For $\vec a=-10\hat j$ and $\vec v=80\hat i+(60-10t)\hat j$, when is the velocity perpendicular to the acceleration?

A. At $t=6\text{ s}$, when the projectile reaches its highest point

B. At $t=0\text{ s}$, before the projectile begins its motion

C. At $t=12\text{ s}$, when the projectile returns to its launch level

D. At $t=8\text{ s}$, when the vertical velocity becomes negative

Answer: (A)

Explanation

Perpendicular vectors have zero dot product, and here $\vec v\cdot\vec a=-10(60-10t)$, which vanishes at $t=6\text{ s}$. At that instant the vertical velocity is zero, but the horizontal velocity remains nonzero, so the velocity itself is not zero.

16. For $\vec u=80\hat i+60\hat j$ and $\vec v=80\hat i+(60-10t)\hat j$, at what mathematical time are the velocity and initial velocity perpendicular?

A. $t=\frac{10}{3}\text{ s}$, during the ascending part of flight

B. $t=\frac{50}{3}\text{ s}$, although this lies beyond the 12-second flight

C. $t=6\text{ s}$, at the projectile's highest point

D. $t=12\text{ s}$, when the projectile reaches its landing height

Answer: (B)

Explanation

Setting $\vec v\cdot\vec u=0$ gives $6400+60(60-10t)=0$ and therefore $t=\frac{50}{3}\text{ s}$. This mathematical solution is not physically realized because the projectile has already completed its 12-second flight.

17. When a projectile lands at the same height from which it was launched, how is its final velocity related to its launch velocity?

A. $\vec v_f=u\cos\theta\,\hat i-u\sin\theta\,\hat j$

B. $\vec v_f=u\sin\theta\,\hat i-u\cos\theta\,\hat j$

C. $\vec v_f=u\cos\theta\,\hat i+u\sin\theta\,\hat j$

D. $\vec v_f=-u\cos\theta\,\hat i+u\sin\theta\,\hat j$

Answer: (A)

Explanation

At equal launch and landing heights, the horizontal component is preserved and the vertical component reverses sign, giving the stated final velocity. The unchanged positive vertical component describes the launch velocity rather than the returning velocity.

18. Which vector relation defines the torque of a force about a point?

A. $\vec\tau=\vec F\cdot\vec p$

B. $\vec\tau=\vec r\times\vec F$

C. $\vec\tau=\vec v\times\vec p$

D. $\vec\tau=\vec r\cdot\vec p$

Answer: (B)

Explanation

Torque about a point is defined as the cross product of the position vector from that point with the applied force. Expressions involving momentum describe angular-momentum-related quantities rather than the definition of torque.

19. Starting from a position vector, what sequence of operations produces velocity, acceleration, and force?

A. Differentiate once, differentiate twice, then multiply acceleration by mass

B. Differentiate once, integrate once, then multiply velocity by mass

C. Integrate twice, differentiate once, then divide velocity by mass

D. Integrate once, integrate twice, then divide acceleration by mass

Answer: (A)

Explanation

The first derivative of position is velocity, the second derivative is acceleration, and Newton's second law gives force as mass times acceleration. Integration reverses differentiation and therefore does not produce this sequence from position.

20. Which equation directly relates the horizontal coordinate and vertical coordinate of a ground-level projectile’s path?

A. $x=u_x t$

B. $v_y=u_y-gt$

C. $y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}$

D. $y=u_y t-\frac{1}{2}gt^2$

Answer: (C)

Explanation

The trajectory equation gives vertical position directly as a function of horizontal position. The other equations relate position or velocity to time rather than relating $x$ and $y$ directly.

21. How can the horizontal range be obtained from a projectile’s trajectory equation?

A. Set $y$ equal to its maximum value and solve for time.

B. Set $y=0$ and select the nonzero value of $x$.

C. Set $\frac{dy}{dx}=0$ and use the resulting value of $y$.

D. Set $x=0$ and select the larger value of $y$.

Answer: (B)

Explanation

The landing points lie where the projectile returns to ground, so setting $y=0$ gives the roots and the nonzero root is the range. The condition $\frac{dy}{dx}=0$ instead identifies the horizontal location of maximum height.

22. For the trajectory $y=x-\frac{x^2}{80}$, what is the horizontal range?

A. $80\ \mathrm{m}$

B. $20\ \mathrm{m}$

C. $160\ \mathrm{m}$

D. $40\ \mathrm{m}$

Answer: (A)

Explanation

Setting $y=0$ gives $x-\frac{x^2}{80}=0$, whose roots are $x=0$ and $x=80\ \mathrm{m}$. The nonzero root represents the landing range, whereas the zero root is the launch point.

23. What procedure identifies the maximum height from a projectile trajectory equation written as $y=f(x)$?

A. Set $y=0$, then use the larger root as the maximum height.

B. Set the horizontal velocity equal to zero, then solve for the vertical coordinate.

C. Set $\frac{dy}{dx}=0$, then substitute that $x$ into the trajectory equation.

D. Set $\frac{d^2y}{dx^2}=0$, then substitute that value into the equation.

Answer: (C)

Explanation

At the highest point, the slope of the trajectory is zero, so solving $\frac{dy}{dx}=0$ locates the corresponding $x$ and substitution gives the height. Setting $y=0$ finds ground intersections rather than the maximum.

24. For a projectile launched and landing at the same level with fixed initial speed $u$, which launch angle gives the greatest range?

A. $45^\circ$, with maximum range $\frac{u^2}{g}$

B. $60^\circ$, with maximum range $\frac{2u^2}{g}$

C. $90^\circ$, with maximum range $\frac{u^2}{g}$

D. $30^\circ$, with maximum range $\frac{u^2}{2g}$

Answer: (A)

Explanation

For ground-to-ground motion, the range reaches its maximum at $45^\circ$ and equals $\frac{u^2}{g}$. A vertical launch has no horizontal displacement, so it cannot produce the greatest range.

25. A projectile is launched and lands at the same height with fixed speed. Which pair of angles produces the same horizontal range?

A. $20^\circ$ and $40^\circ$

B. $30^\circ$ and $30^\circ$

C. $45^\circ$ and $60^\circ$

D. $30^\circ$ and $60^\circ$

Answer: (D)

Explanation

Complementary angles have equal ranges because their doubled angles have equal sine values. Equal launch angles describe the same trajectory rather than forming the distinct complementary-angle pair tested here.

26. What is the standard strategy for solving projectile motion when no horizontal force acts?

A. Use one constant acceleration for both directions and solve for a single displacement.

B. Resolve the initial velocity into components, solve horizontal and vertical motion separately, then recombine the results.

C. Solve the horizontal motion first and assume the vertical velocity remains unchanged.

D. Treat the entire motion as vertical free fall and ignore the horizontal component.

Answer: (B)

Explanation

Projectile motion is handled by resolving the initial velocity into horizontal and vertical components and applying the appropriate equations to each direction. The horizontal component remains constant in this situation, while gravity changes the vertical velocity.

27. For a projectile launched from ground level and landing at ground level with speed $u$ and angle $\theta$, which set of relations gives its flight time, maximum height, and range?

A. $T=\frac{2u\cos\theta}{g},\quad H=\frac{u^2\cos^2\theta}{2g},\quad R=\frac{u^2\cos 2\theta}{g}$

B. $T=\frac{u^2\sin\theta}{g},\quad H=\frac{u\sin^2\theta}{2g},\quad R=\frac{2u\sin 2\theta}{g}$

C. $T=\frac{2u\sin\theta}{g},\quad H=\frac{u^2\sin^2\theta}{2g},\quad R=\frac{u^2\sin 2\theta}{g}$

D. $T=\frac{u\sin\theta}{g},\quad H=\frac{2u^2\sin^2\theta}{g},\quad R=\frac{2u^2\sin 2\theta}{g}$

Answer: (C)

Explanation

These are the standard ground-to-ground projectile relations for time of flight, maximum height, and horizontal range. They do not generally apply directly when the projectile is launched from a tower because the launch and landing heights differ.

28. When horizontal acceleration is zero and the initial components are $u_x$ and $u_y$, which component-based relations are valid?

A. $T=\frac{2u_x}{g},\quad H=\frac{u_x^2}{2g},\quad R=\frac{2u_xu_y}{g}$

B. $T=\frac{u_y}{2g},\quad H=\frac{2u_y^2}{g},\quad R=\frac{u_xu_y}{2g}$

C. $T=\frac{2u_y}{g},\quad H=\frac{u_y^2}{2g},\quad R=\frac{2u_xu_y}{g}$

D. $T=\frac{2u_y}{g},\quad H=\frac{u_x^2}{2g},\quad R=\frac{u_x^2+u_y^2}{g}$

Answer: (C)

Explanation

With zero horizontal acceleration, the vertical component determines flight time and height, while the product of the horizontal and vertical components determines range. If horizontal acceleration were present, the range would require the general horizontal displacement equation.

29. Why is the range greatest at a launch angle of $45^\circ$ for fixed-speed ground-to-ground motion?

A. Because the horizontal and vertical displacements cancel at that angle.

B. Because the flight time is smallest at that angle.

C. Because the vertical component of velocity becomes zero at that angle.

D. Because $\sin 2\theta$ reaches its maximum value of 1 at that angle.

Answer: (D)

Explanation

The range formula contains the factor $\sin 2\theta$, which is largest when $2\theta=90^\circ$, giving $\theta=45^\circ$. The vertical component is not zero at this angle, and the maximum range does not result from minimizing flight time.

30. Why do complementary angles produce the same ground-to-ground range for a fixed launch speed?

A. Their vertical components are equal because $\sin\theta=\sin(90^\circ-\theta)$.

B. Their range factors are equal because $\sin(180^\circ-2\theta)=\sin 2\theta$.

C. Their horizontal components are equal because $\cos\theta=\cos(90^\circ-\theta)$.

D. Their flight times are equal because both angles have the same tangent.

Answer: (B)

Explanation

For complementary angles, the doubled angles are supplementary, and supplementary angles have equal sine values, producing equal values of $R=\frac{u^2\sin 2\theta}{g}$. Their individual horizontal and vertical components are generally exchanged rather than equal.

31. Two projectiles reach the same maximum height but have different launch angles. What can be concluded about their motion?

A. They have equal vertical launch components and equal flight times

B. They have equal ranges because their maximum heights match

C. They have equal horizontal launch components and equal ranges

D. They have equal launch speeds and equal horizontal components

Answer: (A)

Explanation

Equal maximum heights require equal initial vertical components, and those components determine equal flight times. Matching maximum height does not determine equal horizontal range because range also depends on the horizontal component.

32. A projectile has a specified horizontal range equal to its specified maximum height. What launch-angle condition follows?

A. The launch angle satisfies $\tan\theta=4$

B. The launch angle satisfies $\tan\theta=2$

C. The launch angle satisfies $\tan\theta=\frac{1}{2}$

D. The launch angle satisfies $\tan\theta=\frac{1}{4}$

Answer: (A)

Explanation

For a projectile whose range equals its maximum height, substituting the standard range and height expressions gives $\tan\theta=4$. The other tangent values do not satisfy that range-to-height relationship.

33. Which procedure correctly determines whether a projectile's motion is one-dimensional or two-dimensional?

A. Resolve velocity into components, calculate $\tan\alpha=\frac{v_y}{v_x}$, and check whether the direction angle stays constant

B. Calculate the maximum height and decide whether the vertical displacement exceeds the horizontal displacement

C. Find the total distance traveled and check whether the acceleration magnitude changes

D. Compare the launch speed with gravity and check whether the range remains constant

Answer: (A)

Explanation

The motion's dimensional character is determined by its velocity components and whether the direction angle remains constant. Comparing unrelated quantities such as speed, height, or total distance does not establish the number of motion dimensions.

34. A projectile reaches its original landing point, after which it continues over a lower level under changed gravity. How is the new range related to the original range and the extra distance?

A. It is $R'=2R+x$, where $x$ is the added horizontal distance

B. It is $R'=\frac{R}{2}+x$, where $x$ is the added horizontal distance

C. It is $R'=R+x$, where $x$ is the added horizontal distance

D. It is $R'=\frac{R+x}{2}$, where $x$ is the added horizontal distance

Answer: (B)

Explanation

In this setup, the new range is expressed as the original contribution plus the additional horizontal travel, giving $R'=\frac{R}{2}+x$. Treating the original contribution as the full $R$ or doubling it misrepresents the stated geometry and timing.

35. How should the additional horizontal distance be calculated after the projectile enters a region with gravitational acceleration $g'$?

A. Use $x=u\cos\theta\,\frac{2h_{\max}}{g}$

B. Use $x=u\sin\theta\,\frac{2h_{\max}}{g'}$

C. Use $x=u\sin\theta\,\frac{2h_{\max}}{g}$

D. Use $x=u\cos\theta\,\frac{2h_{\max}}{g'}$

Answer: (D)

Explanation

The additional distance equals the horizontal velocity component multiplied by the return time under the changed gravity, so $x=u\cos\theta\,\frac{2h_{\max}}{g'}$. The original gravity belongs to the original trajectory, while the changed gravity controls this additional descent.

36. What is the original range of a projectile launched with speed $u$ at angle $\theta$ under gravitational acceleration $g$?

A. $R=\frac{u\sin 2\theta}{g^2}$

B. $R=\frac{u^2\cos 2\theta}{g}$

C. $R=\frac{u^2\sin\theta}{2g}$

D. $R=\frac{u^2\sin 2\theta}{g}$

Answer: (D)

Explanation

For launch and landing at the same level, the original range is $R=\frac{u^2\sin 2\theta}{g}$. Replacing the squared speed, the sine of the double angle, or the gravitational factor changes the standard range relationship.

37. What coordinate choice best simplifies projectile motion on an inclined plane?

A. Choose axes parallel and perpendicular to the plane, then resolve velocity and gravity along them

B. Choose radial and tangential axes, then treat gravity as perpendicular to the plane

C. Choose horizontal and vertical axes, then resolve gravity into equal components

D. Choose axes along the launch direction and resolve gravity into horizontal components

Answer: (A)

Explanation

Axes parallel and perpendicular to the incline align the problem with the surface and allow both the initial velocity and gravity to be resolved naturally. The other coordinate choices do not directly separate motion relative to the plane.

38. For an inclined plane at angle $\theta$, which gravitational components are correct?

A. $g_{\perp}=g\cos\theta$ and $g_{\parallel}=g\sin\theta$

B. $g_{\perp}=g\sin\theta$ and $g_{\parallel}=g\cos\theta$

C. $g_{\perp}=g\tan\theta$ and $g_{\parallel}=g\cos\theta$

D. $g_{\perp}=g\sin 2\theta$ and $g_{\parallel}=g\cos 2\theta$

Answer: (A)

Explanation

Resolving gravity relative to the incline gives $g_{\perp}=g\cos\theta$ and $g_{\parallel}=g\sin\theta$. Interchanging sine and cosine assigns the components to the wrong directions.

39. After resolving a projectile's motion relative to an inclined plane, which roles do the two directions have?

A. Both directions determine flight time independently, while distance follows from launch speed

B. Along-plane motion determines flight time, while perpendicular motion determines distance

C. Perpendicular motion determines launch speed, while along-plane motion determines gravity

D. Perpendicular motion determines flight time, while along-plane motion determines distance

Answer: (D)

Explanation

Once the components are resolved, the two directions are treated as independent one-dimensional motions: the perpendicular component controls the time in the air, and the along-plane component controls the distance. The reversed assignment confuses the roles of the two axes.

40. A particle is dropped from rest from a tower of height $h$. Which expressions give its time to reach the ground and its final speed, respectively?

A. $t=\sqrt{\frac{2h}{g}}$ and $v=\sqrt{2gh}$

B. $t=\frac{2h}{g}$ and $v=\frac{2gh}{g}$

C. $t=\sqrt{\frac{2g}{h}}$ and $v=\sqrt{\frac{h}{2g}}$

D. $t=\sqrt{\frac{h}{2g}}$ and $v=\sqrt{\frac{2h}{g}}$

Answer: (A)

Explanation

For an object released from rest, constant-acceleration motion gives the flight time $t=\sqrt{\frac{2h}{g}}$ and final speed $v=\sqrt{2gh}$. A thrown object would require a nonzero initial velocity, but a dropped object starts from rest.

41. A ball is thrown horizontally from a tower while another ball is dropped from the same height at the same instant. How do their vertical motions compare, ignoring air resistance?

A. The dropped ball lands later because the thrown ball has greater total speed.

B. The thrown ball lands later because its horizontal speed reduces its downward acceleration.

C. They reach the ground together because both have zero initial vertical velocity.

D. They have different vertical accelerations because one ball is moving horizontally.

Answer: (C)

Explanation

Both balls begin with zero vertical velocity and experience the same downward acceleration, so their vertical motion and flight time are identical. The horizontal launch speed changes the horizontal range, not the vertical time of flight.

42. A particle is projected horizontally with speed $u$ and remains in flight for time $t$. What are its horizontal range and downward vertical velocity at that time?

A. $R=\frac{1}{2}gt^2$ and $v_y=u$ downward

B. $R=ut$ and $v_y=gt$ downward

C. $R=ut^2$ and $v_y=\frac{1}{2}gt^2$ downward

D. $R=\frac{1}{2}ut^2$ and $v_y=gt^2$ downward

Answer: (B)

Explanation

Horizontal motion has constant speed, giving range $R=ut$, while vertical motion starts with zero vertical velocity and gains speed $gt$ downward. The expression involving $\frac{1}{2}gt^2$ describes vertical displacement rather than vertical velocity.

43. A package is released from an aircraft moving horizontally at constant velocity. What is the package's initial velocity immediately after release, relative to the ground?

A. It is opposite to the aircraft's velocity because the package begins falling.

B. It points vertically downward with magnitude determined by its height.

C. It is zero because the package is no longer supported by the aircraft.

D. It equals the aircraft's velocity at the instant of release.

Answer: (D)

Explanation

At release, the package retains the velocity of the moving aircraft, so its initial ground-frame velocity matches the aircraft's velocity. Gravity then changes its velocity after release by adding a downward component.

44. An object is at rest relative to a passenger on a train but moving relative to an observer standing beside the track. What principle does this illustrate?

A. An object's velocity is independent of the observer's reference frame.

B. The train's motion cancels the object's velocity in every reference frame.

C. An object at rest for one observer must be at rest for every observer.

D. Motion is described relative to a chosen observer or reference frame.

Answer: (D)

Explanation

Relative motion depends on the observer or reference frame, so the same object can appear stationary to one observer and moving to another. The other statements incorrectly treat motion as independent of the chosen frame.

45. Two objects have velocities measured in the same reference frame. Which equation gives the velocity of A with respect to B?

A. $\vec v_{A/B}=\frac{\vec v_A}{\vec v_B}$

B. $\vec v_{A/B}=\vec v_A-\vec v_B$

C. $\vec v_{A/B}=\vec v_A+\vec v_B$

D. $\vec v_{A/B}=\vec v_B-\vec v_A$

Answer: (B)

Explanation

The velocity observed for A from B is found by subtracting B's velocity from A's velocity, yielding $\vec v_{A/B}=\vec v_A-\vec v_B$. Reversing the subtraction gives the velocity of B relative to A.

46. Object B moves at $10\hat i$ m/s and object A moves at $2\hat i$ m/s in the same frame. What is B's velocity relative to A?

A. $8\hat i$ m/s

B. $-8\hat i$ m/s

C. $12\hat i$ m/s

D. $5\hat i$ m/s

Answer: (A)

Explanation

Because B is observed relative to A, subtract A's velocity from B's: $\vec v_{B/A}=10\hat i-2\hat i=8\hat i$ m/s. The negative result would describe A's velocity relative to B instead.

47. Which sequence is the appropriate general method for solving a relative-motion problem?

A. Find each object's speed first, add the speeds, and assign the direction from the faster object.

B. Identify observer and object, write the relative equation, substitute components, then find magnitude and direction.

C. Choose a direction first, ignore the observer, and use the larger velocity as the relative velocity.

D. Calculate the distance traveled by each object, compare their times, and omit vector components.

Answer: (B)

Explanation

A relative-motion solution begins by identifying who is observed from whom, then uses the corresponding vector equation before resolving components and finding magnitude or direction. Adding speeds without considering vector directions and observer order can produce an incorrect result.

48. A bird moves at $-10\sqrt3\hat j$ m/s while a man moves at $10\hat i$ m/s. What is the bird's velocity relative to the man?

A. $-10\hat i-10\sqrt3\hat j$ m/s, with magnitude $20$ m/s and direction $60^\circ$ south of west

B. $-10\hat i+10\sqrt3\hat j$ m/s, with magnitude $20$ m/s and direction $60^\circ$ north of west

C. $-10\hat i-10\sqrt3\hat j$ m/s, with magnitude $10\sqrt3$ m/s and direction $30^\circ$ south of west

D. $10\hat i-10\sqrt3\hat j$ m/s, with magnitude $20$ m/s and direction $60^\circ$ south of east

Answer: (A)

Explanation

Subtracting the man's velocity from the bird's gives $\vec v_{bird/man}=-10\hat i-10\sqrt3\hat j$ m/s. Its components produce magnitude $20$ m/s and an angle of $60^\circ$ south of west; changing either component changes the direction or magnitude.

49. A flag is carried by a man walking through moving air. In which direction does the flag point?

A. Along the man's velocity relative to the wind, $\vec v_{man/wind}=\vec v_{man}-\vec v_{wind}$.

B. Along the sum of the wind's and man's ground velocities.

C. Along the wind's velocity relative to the man, $\vec v_{wind/man}=\vec v_{wind}-\vec v_{man}$.

D. Opposite to the wind's ground velocity, regardless of the man's motion.

Answer: (C)

Explanation

The flag responds to the air motion seen from the man, represented by $\vec v_{wind/man}=\vec v_{wind}-\vec v_{man}$. The man's motion therefore changes the apparent wind direction, so the ground-frame wind direction alone is insufficient.

50. A person walking in rain wants to choose the direction for holding an umbrella. Which velocity should determine the umbrella's orientation?

A. The person's velocity relative to the rain, with the umbrella pointed along the apparent rain motion.

B. The rain's velocity relative to the ground, with the umbrella aligned with the person's walking direction.

C. The rain's velocity relative to the person, with the umbrella directed against the apparent incoming rain.

D. The vector sum of the two ground speeds, with the umbrella held vertically upward.

Answer: (C)

Explanation

The person experiences the rain through its relative velocity, $\vec v_{rain/man}=\vec v_{rain}-\vec v_{man}$, so the umbrella should oppose that apparent incoming direction. Using the rain's ground velocity ignores how the person's motion changes the observed direction.

51. A man runs horizontally at $8\hat i\,\text{m/s}$ while rain has velocity $8\hat i\,\text{m/s}$ relative to the ground. How does the rain appear to the man?

A. It moves horizontally forward

B. It moves horizontally backward

C. It remains stationary in the air

D. It falls vertically downward

Answer: (D)

Explanation

The rain’s horizontal velocity relative to the man is $8\hat i-8\hat i=0$, so only its downward component remains apparent. A horizontal apparent motion would require unequal horizontal velocities.

52. An object is projected horizontally from height $h$ with speed $u$. Which expression gives its horizontal range?

A. $u\sqrt{\frac{2g}{h}}$

B. $u\sqrt{\frac{2h}{g}}$

C. $h\sqrt{\frac{2u}{g}}$

D. $\frac{h}{u}\sqrt{\frac{2}{g}}$

Answer: (B)

Explanation

The fall time is $\sqrt{\frac{2h}{g}}$, and multiplying it by the horizontal speed $u$ gives the range. The fall time itself does not depend on the horizontal speed.

53. An aeroplane releases a particle from a height of $320\,\text{m}$ while moving horizontally at $40\,\text{m/s}$. Taking $g=10\,\text{m/s}^2$, where is the particle horizontally when it reaches the ground?

A. $160\,\text{m}$ from the release point

B. $320\,\text{m}$ from the release point

C. $80\,\text{m}$ from the release point

D. $400\,\text{m}$ from the release point

Answer: (B)

Explanation

The fall time is $t=\sqrt{\frac{2(320)}{10}}=8\,\text{s}$, so the horizontal distance is $R=40\times8=320\,\text{m}$. The height determines the time, while the plane’s horizontal speed determines the range.

54. How does a particle dropped from a moving aeroplane appear to move in the aeroplane’s frame and in the ground frame?

A. Vertically in both frames because gravity acts downward

B. Parabolically in the aeroplane frame and vertically in the ground frame

C. Vertically in the aeroplane frame and parabolically in the ground frame

D. Horizontally in the aeroplane frame and circularly in the ground frame

Answer: (C)

Explanation

The particle retains the aeroplane’s horizontal velocity, so its relative horizontal velocity is zero and it falls vertically relative to the aeroplane. Relative to the ground, its horizontal motion combines with vertical free fall to produce a parabola.

55. Two objects move in the same direction with velocities $v_A$ and $v_B$. What is the velocity of A relative to B?

A. $v_A-v_B$

B. $v_B-v_A$

C. $\frac{v_A}{v_B}$

D. $v_A+v_B$

Answer: (A)

Explanation

Relative velocity is found by subtracting the observer’s velocity from the object’s velocity, giving $v_{A/B}=v_A-v_B$. Reversing the order gives the velocity of B relative to A, with the opposite sign.

56. Object A travels at $40\,\text{m/s}$ and object B at $10\,\text{m/s}$ in the same direction, with A initially $90\,\text{m}$ behind B. How long does A take to catch B?

A. $9\,\text{s}$

B. $2\,\text{s}$

C. $3\,\text{s}$

D. $6\,\text{s}$

Answer: (C)

Explanation

A closes the gap at the relative speed $40-10=30\,\text{m/s}$, so the catch-up time is $90/30=3\,\text{s}$. Using the sum of the speeds would incorrectly describe motion in opposite directions.

57. Two accelerated objects meet after time $t$, with A initially ahead of B by distance $d$. Which displacement relation must hold at the meeting time?

A. $x_A=x_B+d$

B. $x_A=x_B-d$

C. $x_A+x_B=d$

D. $x_A=d\,x_B$

Answer: (A)

Explanation

A begins a distance $d$ ahead, so A’s displacement must exceed B’s displacement by that initial separation: $x_A=x_B+d$. The relation can then be combined with $x=ut+\frac{1}{2}at^2$ to determine the meeting time.

58. Two projectiles are launched simultaneously from the same height and experience the same downward gravitational acceleration. What is their relative acceleration?

A. Equal to $g$ downward

B. Zero

C. Directed along their relative velocity

D. Equal to $2g$ downward

Answer: (B)

Explanation

Their gravitational accelerations are identical, so $a_{A/B}=a_A-a_B=0$. The common gravitational acceleration cancels in the relative description.

59. Two projectiles are launched simultaneously from the same vertical level. Which condition is required for them to collide in the air?

A. $u_1\cos\theta_1+u_2\cos\theta_2=0$

B. $u_1\sin\theta_1+u_2\sin\theta_2=g$

C. $u_1\sin\theta_1=u_2\sin\theta_2$

D. $u_1\cos\theta_1=u_2\cos\theta_2$

Answer: (C)

Explanation

Equal initial vertical components ensure that the projectiles have the same vertical position at every common time, allowing a collision to be determined by their horizontal motion. Equal horizontal components are not the required condition.

60. How does one airborne projectile appear to move relative to another when both have the same gravitational acceleration?

A. Along a vertical line with relative acceleration $g$

B. Along a straight line at constant relative velocity

C. Along a circular path with constant relative speed

D. Along a parabola with increasing relative speed

Answer: (B)

Explanation

The equal gravitational accelerations cancel, leaving zero relative acceleration and therefore constant relative velocity. Constant relative velocity produces a straight-line relative path, even though each projectile follows a parabola in the ground frame.

61. Two projectiles are launched simultaneously from the same level in opposite horizontal directions, with ranges $R_1$ and $R_2$. What condition permits an air collision when their initial horizontal separation is $x$?

A. $x\ge R_1+R_2$

B. $x=R_1R_2$

C. $x\le |R_1-R_2|$

D. $x\le R_1+R_2$

Answer: (D)

Explanation

Moving in opposite horizontal directions causes the projectiles’ traveled horizontal distances to add, so an air collision requires $x\le R_1+R_2$. Equality corresponds to meeting at ground level; a difference of ranges applies to the same-direction situation.

62. What does the velocity of a river relative to the ground represent?

A. The velocity measured by a stationary ground observer

B. The velocity of the river relative to a moving boat

C. The velocity of the swimmer relative to the ground

D. The velocity measured by a swimmer relative to the river

Answer: (A)

Explanation

River velocity relative to the ground is defined from the viewpoint of an observer stationary on the ground. A swimmer’s or boat’s motion introduces a different reference frame and therefore a different relative velocity.