Flashcards: Projectile Motion Fundamentals — 79 cards

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1Question

What type of motion is projectile motion?

Answer

Two-dimensional motion along independent horizontal and vertical axes.

2Question

What can be ignored when analyzing horizontal motion in projectile motion?

Answer

Vertical motion can be ignored.

3Question

What can be ignored when analyzing vertical motion in projectile motion?

Answer

Horizontal motion can be ignored.

4Question

When is air resistance considered in the standard projectile model?

Answer

Only if the problem explicitly includes it.

5Question

What value is typically used for gravitational acceleration in projectile problems?

Answer

9.8 m/s² or approximately 10 m/s² unless specified otherwise.

6Question

What is the direction and magnitude of acceleration for a particle under gravity alone?

Answer

Vertically downward with magnitude g.

7Question

What is the projectile acceleration vector with upward and rightward positive?

Answer

a⃗=−gj^\vec a = -g\hat j

8Question

What is the horizontal acceleration of a projectile in standard conditions?

Answer

Zero.

9Question

Why does the horizontal velocity remain constant during projectile flight?

Answer

Because the horizontal acceleration is zero.

10Question

What are the initial velocity components of a particle projected at 53° to vertical at 100 m/s?

Answer

ux=80 m s−1u_x=80\,\mathrm{m\,s^{-1}} and uy=60 m s−1u_y=60\,\mathrm{m\,s^{-1}}.

11Question

What is the velocity vector v⃗(t)\vec v(t) at time t for the example with g=10 m s−2g=10\,\mathrm{m\,s^{-2}}?

Answer

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

12Question

What is the position vector r⃗(t)\vec r(t) at time t for the particle projected from the origin?

Answer

r⃗(t)=80ti^+(60t−5t2)j^ m\vec r(t)=80t\hat i+(60t-5t^2)\hat j\,\mathrm{m}.

13Question

How is the time to reach the highest point tmaxt_{\mathrm{max}} calculated?

Answer

By setting vertical velocity to zero: tmax=uygt_{\mathrm{max}}=\frac{u_y}{g}.

14Question

What are the time of flight and range for the particle returning to the same level?

Answer

Time of flight T=12 sT=12\,\mathrm{s} and range R=960 mR=960\,\mathrm{m}.

15Question

How is the maximum height HmaxH_{\mathrm{max}} calculated in the example?

Answer

Hmax=uy22gH_{\mathrm{max}}=\frac{u_y^2}{2g} equals 180 m.

16Question

What is the trajectory equation after eliminating time from x=80tx=80t and y=60t−5t2y=60t-5t^2?

Answer

y=−x21280+3x4y=-\frac{x^2}{1280}+\frac{3x}{4}, a downward-opening parabola.

17Question

What is the initial velocity vector of a projectile launched at speed u and angle θ?

Answer

u⃗=ucos⁡θ i^+usin⁡θ j^\vec{u}=u\cos\theta\,\hat{i}+u\sin\theta\,\hat{j}

18Question

What is the velocity vector of a projectile at time t?

Answer

v⃗=ucos⁡θ i^+(usin⁡θ−gt) j^\vec{v}=u\cos\theta\,\hat{i}+(u\sin\theta-gt)\,\hat{j}

19Question

What are the coordinates of a projectile at time t?

Answer

x=ucos⁡θ tx=u\cos\theta\,t and y=usin⁡θ t−12gt2y=u\sin\theta\,t-\frac{1}{2}gt^2

20Question

What is the trajectory equation of a projectile eliminating time t?

Answer

y=xtan⁡θ−gx22u2cos⁡2θy=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}

21Question

What is the time to reach maximum height for a projectile returning to launch level?

Answer

tup=usin⁡θgt_{\mathrm{up}}=\frac{u\sin\theta}{g}

22Question

What is the total time of flight for a projectile returning to launch level?

Answer

T=2usin⁡θgT=\frac{2u\sin\theta}{g}

23Question

What is the maximum height and range of a projectile returning to launch level?

Answer

Maximum height: H=u2sin⁡2θ2gH=\frac{u^2\sin^2\theta}{2g}; Range: R=u2sin⁡2θgR=\frac{u^2\sin 2\theta}{g}

24Question

When is velocity perpendicular to acceleration in vector terms?

Answer

When their dot product v⃗⋅a⃗=0\vec v\cdot\vec a=0.

25Question

At what time is velocity perpendicular to acceleration for v⃗=80i^+(60−10t)j^\vec v=80\hat i+(60-10t)\hat j and a⃗=−10j^\vec a=-10\hat j?

Answer

At t=6 st=6\text{ s}, the highest point.

26Question

When is velocity perpendicular to initial velocity mathematically?

Answer

When their dot product v⃗⋅u⃗=0\vec v\cdot\vec u=0.

27Question

What is the mathematical time when u⃗=80i^+60j^\vec u=80\hat i+60\hat j and v⃗=80i^+(60−10t)j^\vec v=80\hat i+(60-10t)\hat j are perpendicular?

Answer

At t=503 st=\frac{50}{3}\text{ s}, which is not physically possible if flight ends at 12 s.

28Question

What is the final velocity formula when a projectile lands at original height?

Answer

v⃗f=ucos⁡θ i^−usin⁡θ j^\vec v_f=u\cos\theta\,\hat i - u\sin\theta\,\hat j.

29Question

What is the formula for torque about a point?

Answer

τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}

30Question

What does differentiating a position vector once give?

Answer

Velocity

31Question

What does differentiating a position vector twice give?

Answer

Acceleration

32Question

How is force related to acceleration and mass?

Answer

Force equals mass times acceleration

33Question

How do you find the horizontal range from a trajectory equation?

Answer

Set y=0y=0 and take the nonzero value of xx.

34Question

What horizontal range results from setting y=0y=0 in y=x−x280y=x-\frac{x^2}{80}?

Answer

80 m80\ \mathrm{m}

35Question

How do you find the maximum height from a trajectory equation?

Answer

Set dydx=0\frac{dy}{dx}=0 to find xx, then substitute into the equation.

36Question

At what angle is range maximum for fixed initial speed in ground-to-ground motion?

Answer

45 degrees

37Question

What is the maximum range value for fixed initial speed in ground-to-ground motion?

Answer

u2g\frac{u^2}{g}

38Question

Which launch angles produce the same range for fixed speed and equal launch and landing levels?

Answer

θ\theta and 90∘−θ90^\circ - \theta

39Question

How is projectile motion solved in terms of velocity components?

Answer

By resolving initial velocity into horizontal and vertical components and treating motions separately

40Question

How is the maximum height of a projectile expressed in terms of launch speed and angle?

Answer

H=u2sin⁡2θ2gH=\frac{u^2\sin^2\theta}{2g}

41Question

What formula gives the horizontal range for ground-to-ground projectile motion?

Answer

R=u2sin⁡2θgR=\frac{u^2\sin 2\theta}{g}

42Question

How are the initial velocity components uxu_x and uyu_y defined for projectile motion?

Answer

ux=ucos⁡θu_x=u\cos\theta and uy=usin⁡θu_y=u\sin\theta

43Question

What is the time of flight formula using initial vertical velocity component uyu_y?

Answer

T=2uygT=\frac{2u_y}{g}

44Question

At which launch angle is the range of a projectile maximized for fixed speed?

Answer

At θ=45∘\theta=45^\circ

45Question

Why does the range maximize at θ=45∘\theta=45^\circ for fixed launch speed?

Answer

Because sin⁡2θ\sin 2\theta reaches its maximum value of 1 there

46Question

Why do complementary angles θ\theta and 90∘−θ90^\circ-\theta yield the same projectile range?

Answer

Because sin⁡(180∘−2θ)=sin⁡2θ\sin(180^\circ-2\theta)=\sin 2\theta

47Question

What vertical velocity components do two projectiles with equal max heights share?

Answer

They have equal vertical components u1sin⁡θ1=u2sin⁡θ2u_1\sin\theta_1 = u_2\sin\theta_2.

48Question

On what does the range of two projectiles with equal max heights depend?

Answer

Their ranges depend on the products uxuyu_x u_y.

49Question

What is the launch angle θ\theta if a projectile has the same horizontal range and max height?

Answer

The launch angle satisfies tan⁡θ=4\tan\theta = 4.

50Question

How do you check if motion is one-dimensional or two-dimensional using velocity components?

Answer

Calculate tan⁡α=vy/vx\tan\alpha = v_y / v_x and check if the direction angle remains constant.

51Question

What is the formula for the new projectile range on a lower level?

Answer

R′=R2+xR' = \frac{R}{2}+x

52Question

How is the additional horizontal distance after landing calculated with changed gravity?

Answer

x=ucos⁡θ×2hmax⁡g′x = u \cos \theta \times \frac{2h_{\max}}{g'}

53Question

Which axes are chosen for projectile motion on an inclined plane?

Answer

Axes parallel and perpendicular to the plane.

54Question

How is gravitational acceleration resolved on an inclined plane at angle θ?

Answer

Into g⊥=gcos⁡θg_{\perp}=g\cos\theta perpendicular and g∥=gsin⁡θg_{\parallel}=g\sin\theta along the plane.

55Question

What must be done to initial velocity u at angle θ before applying kinematics on an incline?

Answer

Resolve it into components along the chosen axes.

56Question

How is the position vector of a particle with coordinates x and y expressed?

Answer

r⃗=xi^+yj^\vec r=x\hat i+y\hat j

57Question

What does the course treat projectile motion as part of?

Answer

Kinematics involving motion along two independent axes.

58Question

What formula relates the angle θ\theta of velocity with horizontal at time t?

Answer

tan⁡θ=60−10t80\tan\theta=\frac{60-10t}{80}.

59Question

What is the formula for change in velocity between launch and landing at same height?

Answer

Δv⃗=−2usin⁡θ j^\Delta\vec v = -2u\sin\theta\,\hat j.

60Question

How is torque of gravitational force about origin expressed at position r⃗=xi^+yj^\vec r=x\hat i + y\hat j?

Answer

τ⃗=mgxk^\vec\tau = mgx\hat k.

61Question

What is the angular momentum magnitude at highest point about launch point?

Answer

L=mHucos⁡θL = mH u \cos\theta.

62Question

How do you find velocity from force if the particle starts at rest?

Answer

Integrate force divided by mass

63Question

How do you find position vector from velocity?

Answer

Integrate velocity

64Question

How is velocity obtained from r⃗=103t3i^+5t2j^\vec{r} = \frac{10}{3}t^3\hat{i} + 5t^2\hat{j}?

Answer

By differentiating r⃗\vec{r} with respect to time

65Question

How is velocity at t=1 st=1\ \mathrm{s} found for r⃗=103t3i^+5t2j^\vec{r} = \frac{10}{3}t^3\hat{i} + 5t^2\hat{j}?

Answer

By substituting t=1t=1 into the derivative of r⃗\vec{r}

66Question

At what xx and height does y=x−x280y=x-\frac{x^2}{80} reach its maximum?

Answer

Maximum at x=40 mx=40\ \mathrm{m} and height 20 m20\ \mathrm{m}.

67Question

How do you identify tan⁡θ\tan\theta from a given trajectory equation?

Answer

From the coefficient of xx in the equation.

68Question

How do you identify initial speed uu from a trajectory equation?

Answer

From the coefficient of x2x^2 in the equation.

69Question

What happens to horizontal velocity when a horizontal force acts besides gravity?

Answer

Horizontal velocity is not constant

70Question

Can initial and final vertical speeds be related if no vertical force acts besides gravity?

Answer

Yes, through vertical motion equations

71Question

What are the horizontal and vertical components of velocity for a projectile launched at 50 m/s at 37°?

Answer

40 m/s horizontal and 30 m/s vertical

72Question

What formula expresses the range RR of a projectile with zero horizontal acceleration?

Answer

R=2uxuygR=\frac{2 u_x u_y}{g}.

73Question

Why is the range greater for the projectile with the greater product uxuyu_x u_y?

Answer

Because range RR is proportional to 2uxuy/g2 u_x u_y / g when horizontal acceleration is zero.

74Question

What type of motion results from zero initial velocity and constant acceleration?

Answer

One-dimensional straight-line motion.

75Question

How is velocity direction relative to the x-axis determined from position vector r⃗(t)\vec r(t)?

Answer

Differentiate r⃗(t)\vec r(t) to get velocity, then use tan⁡α=vy/vx\tan\alpha = v_y / v_x for direction.

76Question

What is the new-to-old range ratio when gravity changes to g′=g81g' = \frac{g}{81}?

Answer

95

77Question

How is the inclined-plane projectile motion problem treated after resolving components?

Answer

As two independent one-dimensional motions.

78Question

Which motion component determines flight time on an inclined plane?

Answer

The perpendicular motion component.

79Question

Which motion component determines distance along the inclined plane?

Answer

The motion along the plane.

Test yourself with the quiz

Test your knowledge with 62 questions on Projectile Motion Fundamentals.

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

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

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