Study sheet: Projectile Motion Fundamentals

📋 Course Outline

1. Two-Dimensional Motion Framework
2. Projectile Components and Assumptions
3. Projectile Motion Calculations
4. Trajectory and Standard Quantities
5. Velocity, Momentum, and Torque
6. Angular Momentum and Torque Methods
7. Trajectory Equation Applications
8. Projectile Motion Strategies and Formulas
9. Projectile Motion Core Relations
10. Projectile Comparisons and Motion Types
11. Projectile Range with Changed Gravity
12. Projectile Motion on an Inclined Plane
13. Tower Drops and Relative Motion
14. Relative Velocity Fundamentals
15. Relative Motion Applications
16. Projectile and Catch-Up Problems
17. Relative Motion Along a Line
18. Relative Motion of Projectiles

📖 1. Two-Dimensional Motion Framework

★ Must-know

📌 Projectile motion is two-dimensional motion in which the particle moves simultaneously along independent horizontal and vertical axes.

📌 When analyzing horizontal motion, the vertical motion can be ignored, and when analyzing vertical motion, the horizontal motion can be ignored.

Further detail

📌 The position vector of a particle with coordinates x and y relative to the origin is $\vec r=x\hat i+y\hat j$.

• The course treats projectile motion as a part of kinematics involving motion along two independent axes.

💡 Memory Hook

x moves independently, whereas y responds to gravity

📖 2. Projectile Components and Assumptions

📝 Essential Points

• In the standard projectile model, air resistance is neglected unless the problem explicitly includes it.

• The acceleration due to gravity is taken as constant, with $g=9.8\,\mathrm{m\,s^{-2}}$ or approximately $10\,\mathrm{m\,s^{-2}}$ unless another value is specified.

• For a particle in flight under gravity alone, the acceleration is vertically downward with magnitude g, regardless of whether the particle is rising, falling, or moving at an angle.

📐 Formula — With upward and rightward directions chosen as positive, the projectile acceleration is $\vec a=-g\hat j$, so the horizontal acceleration is zero.

📌 Because the horizontal acceleration is zero, the horizontal velocity remains constant throughout the flight before impact.

💡 Memory Hook

No horizontal force → zero horizontal acceleration → constant horizontal velocity

📖 3. Projectile Motion Calculations

★ Must-know

• A particle projected at 100 m/s at 53° to the vertical has an angle of 37° to the horizontal and initial components $u_x=100\cos37^\circ=80\,\mathrm{m\,s^{-1}}$ and $u_y=100\sin37^\circ=60\,\mathrm{m\,s^{-1}}$.

📐 Formula — For this example, the velocity at time t is $\vec v(t)=80\hat i+(60-10t)\hat j\,\mathrm{m\,s^{-1}}$ when $g=10\,\mathrm{m\,s^{-2}}$.

📐 Formula — For the same example, the position vector at time t is $\vec r(t)=80t\hat i+(60t-5t^2)\hat j\,\mathrm{m}$ when the particle is projected from the origin.

📐 Formula — The time to reach the highest point is found by setting the vertical velocity to zero, giving $t_{\mathrm{max}}=\frac{u_y}{g}=\frac{60}{10}=6\,\mathrm{s}$.

📐 Formula — For the example returning to the same level, the time of flight is $T=2t_{\mathrm{max}}=12\,\mathrm{s}$ and the range is $R=u_xT=80\times12=960\,\mathrm{m}$.

📐 Formula — The maximum height in the example is obtained from vertical motion and equals $H_{\mathrm{max}}=\frac{u_y^2}{2g}=\frac{60^2}{2\times10}=180\,\mathrm{m}$.

📐 Formula — Eliminating time from $x=80t$ and $y=60t-5t^2$ gives the trajectory equation $y=-\frac{x^2}{1280}+\frac{3x}{4}$, which is a downward-opening parabola.

Further detail

📐 Formula — At time t, the angle θ made by the velocity with the horizontal satisfies $\tan\theta=\frac{v_y}{v_x}=\frac{60-10t}{80}$.

📐 Formula — At t=2 s, the particle has velocity $\vec v=80\hat i+40\hat j\,\mathrm{m\,s^{-1}}$ and position vector $\vec r=160\hat i+100\hat j\,\mathrm{m}$.

📐 Formula — The angular momentum of the particle about the projection point is calculated using $\vec L=\vec r\times m\vec v$, and the torque of gravity about that point is calculated using $\vec\tau=\vec r\times\vec F$ with $\vec F=m\vec g$.

💡 Memory Hook

Resolve → evolve in time → find position → eliminate time

📖 4. Trajectory and Standard Quantities

📝 Essential Points

📐 Formula — For a projectile launched with speed u at angle θ, the initial velocity is $\vec{u}=u\cos\theta\,\hat{i}+u\sin\theta\,\hat{j}$ and the acceleration is $\vec{a}=-g\hat{j}$.

📐 Formula — At time t, the projectile velocity is $\vec{v}=u\cos\theta\,\hat{i}+(u\sin\theta-gt)\,\hat{j}$.

📐 Formula — The coordinates of the projectile at time t are $x=u\cos\theta\,t$ and $y=u\sin\theta\,t-\frac{1}{2}gt^2$.

📐 Formula — Eliminating time between the coordinate equations gives the trajectory equation $y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}$.

📐 Formula — For a projectile returning to its launch level, the time to reach maximum height is $t_{\mathrm{up}}=\frac{u\sin\theta}{g}$ and the total time of flight is $T=\frac{2u\sin\theta}{g}$.

📐 Formula — For a projectile returning to its launch level, the maximum height is $H=\frac{u^2\sin^2\theta}{2g}$ and the range is $R=\frac{u^2\sin 2\theta}{g}$.

💡 Memory Hook

Resolve velocity → eliminate time → obtain trajectory

📖 5. Velocity, Momentum, and Torque

★ Must-know

📌 Velocity is perpendicular to acceleration when their dot product is zero, so $\vec v\cdot\vec a=0$; for the given acceleration $-10\hat j$ and velocity $80\hat i+(60-10t)\hat j$, this occurs at $t=6\text{ s}$, the highest point.

📌 Velocity is perpendicular to the initial velocity when $\vec v\cdot\vec u=0$; for $\vec u=80\hat i+60\hat j$ and $\vec v=80\hat i+(60-10t)\hat j$, the mathematical time is $t=\frac{50}{3}\text{ s}$, which is not physically possible if the flight ends at 12 s.

📐 Formula — When the projectile lands at its original height, its final velocity is $\vec v_f=u\cos\theta\,\hat i-u\sin\theta\,\hat j$, obtained by reversing the vertical component while preserving the horizontal component.

Further detail

📐 Formula — The change in velocity between launch and landing at the same height is $\Delta\vec v=\vec v_f-\vec v_i=-2u\sin\theta\,\hat j$, and the change in momentum is $\Delta\vec p=m\Delta\vec v$.

📐 Formula — The torque of gravitational force about the origin at position $\vec r=x\hat i+y\hat j$ is $\vec\tau=\vec r\times(-mg\hat j)=mgx\hat k$.

📐 Formula — At the highest point, the angular momentum magnitude about the launch point is $L=mH\,u\cos\theta$, because the position has horizontal coordinate $R/2$ and vertical coordinate H while the velocity is horizontal.

💡 Memory Hook

Horizontal motion stays constant in the basic case, whereas vertical motion changes under gravity

📖 6. Angular Momentum and Torque Methods

★ Must-know

📐 Formula — The torque about a point is $\vec\tau=\vec r\times\vec F$.

• For a position vector, differentiating once gives velocity, differentiating twice gives acceleration, and multiplying acceleration by mass gives force.

Further detail

• If force is given and the particle starts from rest at the origin, integrating force divided by mass gives velocity and integrating velocity gives the position vector.

• For $\vec r=\frac{10}{3}t^3\hat i+5t^2\hat j$, differentiating gives velocity, and at $t=1\ \mathrm{s}$ the velocity is obtained by substituting that time into the derivative.

💡 Memory Hook

Position → velocity → acceleration → force → torque

📖 7. Trajectory Equation Applications

★ Must-know

📐 Formula — For a projectile launched from ground level, the trajectory equation is $y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}$.

• To find the horizontal range from a trajectory equation, set $y=0$ and take the nonzero value of x.

• For $y=x-\frac{x^2}{80}$, setting $y=0$ gives a horizontal range of $80\ \mathrm{m}$.

• To find the maximum height from a trajectory equation, set $\frac{dy}{dx}=0$ to locate the highest point and then substitute that x-coordinate into the trajectory equation.

Further detail

• For $y=x-\frac{x^2}{80}$, the maximum occurs at $x=40\ \mathrm{m}$ and the maximum height is $20\ \mathrm{m}$.

• Comparing a given trajectory equation with the standard form identifies $\tan\theta$ from the coefficient of x and identifies $u$ from the coefficient of $x^2$.

💡 Memory Hook

Range: y = 0; highest point: dy/dx = 0

📖 8. Projectile Motion Strategies and Formulas

★ Must-know

📌 For fixed initial speed in ground-to-ground motion, the range is maximum at $\theta=45^\circ$ and has maximum value $\frac{u^2}{g}$.

📌 For a fixed speed and equal launch and landing levels, launch angles $\theta$ and $90^\circ-\theta$ produce the same range.

• Projectile motion is solved by resolving the initial velocity into horizontal and vertical components, treating horizontal motion separately from vertical motion, and recombining the resulting coordinates or velocities.

Further detail

📌 When a horizontal force acts in addition to gravity, horizontal velocity is not constant, but if no additional vertical force acts, the initial and final vertical speeds can still be related through vertical motion.

• For a projectile launched at $50\ \mathrm{m\,s^{-1}}$ at 37° using the 3–4–5 component values, the horizontal and vertical components are respectively $40\ \mathrm{m\,s^{-1}}$ and $30\ \mathrm{m\,s^{-1}}$.

💡 Memory Hook

Resolve components → use independent horizontal and vertical motion → reconstruct the result

📖 9. Projectile Motion Core Relations

★ Must-know

📐 Formula — For a projectile launched with speed $u$ at angle $\theta$ to the horizontal, the time of flight is $T=\frac{2u\sin\theta}{g}$, the maximum height is $H=\frac{u^2\sin^2\theta}{2g}$, and the horizontal range is $R=\frac{u^2\sin 2\theta}{g}$ for ground-to-ground motion.

📐 Formula — If the initial velocity components are $u_x=u\cos\theta$ and $u_y=u\sin\theta$, then $T=\frac{2u_y}{g}$, $H=\frac{u_y^2}{2g}$, and $R=\frac{2u_xu_y}{g}$ when horizontal acceleration is zero.

📌 For fixed launch speed in ground-to-ground motion, the range is maximum at $\theta=45^\circ$ because $\sin 2\theta$ is maximum when it equals 1.

📌 For fixed launch speed, complementary projection angles $\theta$ and $90^\circ-\theta$ produce the same ground-to-ground range because $\sin(180^\circ-2\theta)=\sin 2\theta$.

Further detail

• At the highest point of a projectile, the vertical velocity is zero and the remaining speed is the horizontal component $u\cos\theta$.

💡 Memory Hook

Ground-to-ground formulas work for level ground, whereas tower launches require a different treatment.

📖 10. Projectile Comparisons and Motion Types

★ Must-know

📌 Two projectiles with equal maximum heights have equal vertical components $u_1\sin\theta_1=u_2\sin\theta_2$ and therefore equal flight times, while their ranges depend on the products $u_xu_y$.

📐 Formula — If a projectile has the same horizontal range and maximum height, then its launch angle satisfies $\tan\theta=4$.

• To determine whether motion is one-dimensional or two-dimensional, find the velocity components, calculate $\tan\alpha=v_y/v_x$, and check whether the direction angle remains constant.

Further detail

📐 Formula — For a projectile, the range is greater for the motion with the greater product $u_xu_y$ because $R=\frac{2u_xu_y}{g}$ when horizontal acceleration is zero.

📌 Motion with zero initial velocity and constant acceleration, or with initial velocity and acceleration along the same line, is one-dimensional straight-line motion; non-collinear initial velocity and acceleration produce two-dimensional motion such as projectile motion.

📐 Formula — For a particle with position vector $\vec r(t)$, differentiate once to obtain velocity, differentiate twice to obtain acceleration, and use $\tan\alpha=v_y/v_x$ to determine the velocity direction relative to the x-axis.

💡 Memory Hook

Equal vertical component → equal maximum height and flight time; equal complementary angles → equal range.

📖 11. Projectile Range with Changed Gravity

★ Must-know

📐 Formula — When a projectile lands on a lower level after its gravity changes, its new range is the old range plus the additional horizontal distance traveled after the original landing point: $R' = \frac{R}{2}+x$.

📐 Formula — The additional horizontal distance is $x=u\cos\theta\,\frac{2h_{\max}}{g'}$, where $g'$ is the new gravitational acceleration.

📐 Formula — The original projectile range is $R=\frac{u^2\sin 2\theta}{g}$.

Further detail

• For the example where gravity is changed to $g'=g/81$, the calculated new-to-old range ratio is 95.

💡 Memory Hook

Increased gravity → shorter flight time and a modified range.

📖 12. Projectile Motion on an Inclined Plane

★ Must-know

• For projectile motion on an inclined plane, choose axes parallel and perpendicular to the plane, then resolve both the initial velocity and gravitational acceleration along these axes.

📐 Formula — For an incline at angle θ, the gravitational acceleration components are $g_{\perp}=g\cos\theta$ perpendicular to the plane and $g_{\parallel}=g\sin\theta$ along the plane.

• For an initial speed u at angle θ, the velocity must be resolved into components along the chosen axes before applying one-dimensional kinematics.

Further detail

• After the components are resolved, the inclined-plane problem is treated as two independent one-dimensional motions: the perpendicular motion determines the flight time, while the along-plane motion determines the distance.

💡 Memory Hook

Along the plane changes horizontal motion; perpendicular to the plane changes vertical motion.

📖 13. Tower Drops and Relative Motion

★ Must-know

📐 Formula — For a particle dropped from rest from height h, the time to reach the ground is $t=\sqrt{\frac{2h}{g}}$ and the final speed is $v=\sqrt{2gh}$.

📌 A particle thrown horizontally from a tower has the same vertical motion as a particle dropped from the same height, because its initial vertical velocity is zero and horizontal acceleration is zero.

📐 Formula — For horizontal projection with horizontal speed u and flight time t, the range is $R=ut$, while the vertical velocity at time t is $v_y=gt$ downward when the initial vertical velocity is zero.

📌 If a particle is released from a moving balloon, train, vehicle, lift, or aircraft, it initially retains the velocity of that moving frame at the instant of release.

Further detail

📐 Formula — The release condition from a moving frame is $\vec v_{\text{particle}}-\vec v_{\text{frame}}=0$, so $\vec v_{\text{particle}}=\vec v_{\text{frame}}$ at release.

💡 Memory Hook

Resolve components → solve vertical motion → calculate horizontal range.

📖 14. Relative Velocity Fundamentals

🔑 Key Concepts & Definitions

• Relative motion : Motion described with respect to a specified observer or reference frame, so an object can appear moving to one observer and at rest to another.

★ Must-know

📐 Formula — The velocity of A with respect to B is $\vec v_{A/B}=\vec v_A-\vec v_B$, where the velocities are measured in the same reference frame.

• If B moves at $10\hat i$ m/s and A moves at $2\hat i$ m/s, then B's velocity relative to A is $8\hat i$ m/s.

Further detail

📐 Formula — The acceleration of A with respect to B is $\vec a_{A/B}=\vec a_A-\vec a_B$, obtained by differentiating the relative-velocity relation.

📐 Formula — The position vector of B with respect to A is $\vec r_{B/A}=\vec r_B-\vec r_A$; differentiating it gives $\vec v_{B/A}=\vec v_B-\vec v_A$.

💡 Memory Hook

Ground view versus observer view: the same object may be moving for one observer and at rest for another.

📖 15. Relative Motion Applications

★ Must-know

• 🔄 A relative-motion problem is solved by these steps:

  1. Identify the observed object and the observer
  2. Write the relative-velocity equation
  3. Substitute the vector components
  4. Determine the magnitude and direction if required

• For a bird moving at $-10\sqrt3\hat j$ m/s and a man moving at $10\hat i$ m/s, the bird's velocity relative to the man is $-10\hat i-10\sqrt3\hat j$ m/s, with magnitude $20$ m/s and direction $60^\circ$ south of west.

📌 A flag carried by a moving man points in the direction of the wind's velocity relative to the man, namely $\vec v_{wind/man}=\vec v_{wind}-\vec v_{man}$.

📌 To determine the direction in which a moving man should hold an umbrella, use the rain's velocity relative to the man, $\vec v_{rain/man}=\vec v_{rain}-\vec v_{man}$, and orient the umbrella against the apparent incoming rain.

Further detail

• If rain moves at $10\sqrt3$ m/s south and a man moves at $10$ m/s east, the rain moves relative to him at $-10\hat i-10\sqrt3\hat j$ m/s, so its speed is $20$ m/s and its direction is $60^\circ$ south of west.

💡 Memory Hook

Identify the observer → subtract velocities → draw the relative vector → determine speed and direction.

📖 16. Projectile and Catch-Up Problems

★ Must-know

📌 If rain has horizontal component $8\hat i$ m/s and a man runs at $8\hat i$ m/s, the rain appears to fall vertically downward to the man because the relative horizontal component is zero.

📐 Formula — For an object dropped horizontally from height $h$ with horizontal speed $u$, the time to reach the ground is $t=\sqrt{\frac{2h}{g}}$ and the horizontal range is $R=u\sqrt{\frac{2h}{g}}$.

• A particle dropped from an aeroplane at height $320$ m moving horizontally at $40$ m/s reaches the ground after $8$ s and travels a horizontal range of $320$ m when $g=10$ m/s².

📌 A dropped particle has a parabolic path relative to the ground but a straight vertical path relative to the aeroplane because the particle and aeroplane have the same horizontal velocity.

Further detail

📐 Formula — For a pursuer moving at $40$ m/s behind an object moving at $10$ m/s with an initial separation of $90$ m, the catch-up time is $t=\frac{90}{40-10}=3$ s.

💡 Memory Hook

Equal horizontal velocities cause zero relative horizontal motion; relative displacement then determines the observed path or meeting time.

📖 17. Relative Motion Along a Line

★ Must-know

📐 Formula — For two objects moving in the same direction with constant velocities, the relative velocity of A with respect to B is $v_{A/B}=v_A-v_B$, so the catch-up time equals the initial gap divided by the relative velocity.

• If A moves at 40 m/s, B moves at 10 m/s, and the initial gap is 90 m, A catches B after $t=90/(40-10)=3\ \text{s}$.

📐 Formula — For accelerated objects that meet after time t, their displacements satisfy $x_A=x_B+d$, where d is the initial separation; using $x=ut+\frac{1}{2}at^2$ gives the meeting equation.

📐 Formula — Relative quantities obey $u_{A/B}=u_A-u_B$, $a_{A/B}=a_A-a_B$, and $s_{A/B}=u_{A/B}t+\frac{1}{2}a_{A/B}t^2$.

📐 Formula — For two projectiles launched simultaneously from the same vertical level and moving under the same downward gravitational acceleration, the relative acceleration is zero: $a_{A/B}=a_A-a_B=0$.

Further detail

📌 The displacement method is generally safer for accelerated catch-up problems, whereas the relative-motion method is quicker when the relative acceleration is zero or the relative quantities are easy to determine.

💡 Memory Hook

Ground-frame displacement versus observer-frame relative motion

📖 18. Relative Motion of Projectiles

🔑 Key Concepts & Definitions

• River velocity relative to ground : the velocity observed by a stationary observer standing on the ground
• Man velocity relative to river : the swimmer's or boat's velocity measured with respect to the water, and it satisfies $\vec v_{m/r}=\vec v_{m/g}-\vec v_{r/g}$

📝 Essential Points

📌 When two projectiles are launched simultaneously from the same vertical level, they can collide in the air only if their initial vertical velocity components are equal: $u_1\sin\theta_1=u_2\sin\theta_2$.

• Because the relative acceleration of two airborne projectiles is zero, one projectile has constant velocity relative to the other and therefore appears to move along a straight-line path.

📐 Formula — For two projectiles launched simultaneously from the same level in opposite horizontal directions, the horizontal separation must satisfy $x\le R_1+R_2$ for an air collision, with equality corresponding to collision at ground level.

• For two projectiles launched simultaneously from the same horizontal level to collide, their vertical velocity components must be equal: $u_1\sin\theta_1=u_2\sin\theta_2$.

• If two projectiles have equal vertical components, collision occurs when the horizontal separation is less than the sum of their horizontal ranges, $x<r_1+r_2$; equality means collision at ground level.

• Downstream motion is along the river, so the ground speed is the sum of the swimmer's speed relative to the river and the river speed, whereas upstream motion is against the river, so the ground speed is their difference.

📐 Formula — For a river of speed $u$ and a swimmer speed $v$ relative to the river crossing a width $d$, the times for downstream and upstream travel over the same distance are respectively $t_{down}=\frac{d}{v+u}$ and $t_{up}=\frac{d}{v-u}$.

• To solve an angled river-crossing problem, resolve the swimmer's velocity into components, use the transverse component to determine crossing time, and combine the parallel component with the river velocity to determine drift.

📐 Formula — If a swimmer moves at speed $v$ at angle $\theta$ to the river direction, crosses a river of width $d$, and has transverse component $v\cos\theta$, the crossing time is $t=\frac{d}{v\cos\theta}$.

• For a swimmer whose velocity component along the river is in the same direction as the current, the downstream drift speed is the sum of that component and the river speed; for the opposite direction, the speeds are subtracted.

📐 Formula — To cross the river by the minimum-distance path and reach the point directly opposite the starting point, the swimmer must cancel the river's parallel velocity, so $v\sin\theta=u$ and therefore $\sin\theta=\frac{u}{v}$.

📐 Formula — For minimum-time crossing, the swimmer moves perpendicular to the river with the full speed across the river, so $\theta=0$, $t_{min}=\frac{d}{v}$, and the drift is $ut_{min}=\frac{ud}{v}$.

• For a swimmer moving at speed b at angle θ to the river direction, the cross-river component is $b\sin\theta$ and the along-river component is $b\cos\theta$.

📐 Formula — The river speed is obtained from its drift during minimum-time crossing as $u=\frac{\text{drift}}{\text{time}}$; a drift of 120 metres in 10 minutes gives $u=12\ \text{m/min}$.

💡 Memory Hook

Equal vertical components → equal vertical motion → possible collision

📊 Synthesis Tables

Trajectory equation shortcuts

Question	Condition	Method
Range	y = 0	Use the nonzero x-root
Maximum height	dy/dx = 0	Find x, then substitute into y
Launch angle and speed	Compare coefficients	Match the equation with the standard trajectory form

Key Projectile Quantities

Quantity	Formula	Depends on
Time of flight	$T=\frac{2u_y}{g}$	Vertical component
Maximum height	$H=\frac{u_y^2}{2g}$	Square of vertical component
Range	$R=\frac{2u_xu_y}{g}$	Horizontal–vertical component product