My1–My5 - Projectile motion and trajectories | OCR A-Level Mathematics B (MEI)

Form a projectile's vector equations from its launch and gravity, then use them to find its flight, launch velocity and Cartesian trajectory. Keep the landing surface explicit, including a specified slope, and decide when the idealised model remains useful.

Assumptions and component equations

Can a projectile hit rising ground while it is still rising? A ball launched with velocity $10\mathbf i+15\mathbf j\,\mathrm{m\,s^{-1}}$ , using $g=10\,\mathrm{m\,s^{-2}}$ , has height $15t-5t^2$ . If the ground is $y=x$ and $x=10t$ , it meets the ground after one second, while its vertical velocity is still $5\,\mathrm{m\,s^{-1}}$ upwards. The landing surface can end a flight before its usual highest point.

First build the free-flight model; the final part will explain why that intersection is a genuine first impact.

You already know

Resolve launch speed $U$ at angle $\theta$ above horizontal into $U\cos\theta$ horizontally and $U\sin\theta$ vertically. Both components share one elapsed time.

Model the projectile as a particle with no air resistance or propulsion after release. Horizontal travel is small enough to regard gravity as parallel everywhere; vertical travel is small enough to take its magnitude as constant. These assumptions ignore size, rotation and deformation, and can be poor for light objects with a large surface area.

Projectile under gravity

During idealised free flight, a projectile's only force is gravity. With $\mathbf i$ horizontal and $\mathbf j$ vertically upwards, its acceleration is $\mathbf a=-g\mathbf j$ . In particular, zero upward velocity does not mean zero acceleration.

Use the specified value of $g$ . If none is specified, use $9.8\,\mathrm{m\,s^{-2}}$ ; another planet or an explicitly simplified model requires its own stated value.

Vector equations for a projectile

With starting position $x_0\mathbf i+y_0\mathbf j$ and initial velocity $u_x\mathbf i+u_y\mathbf j$ ,

\begin{aligned} \mathbf v&=u_x\mathbf i+(u_y-gt)\mathbf j\\ \mathbf r&=(x_0+u_xt)\mathbf i+\left(y_0+u_yt-\tfrac12gt^2\right)\mathbf j \end{aligned}

These follow by inserting $\mathbf a=-g\mathbf j$ into the supplied vector constant-acceleration equations. These particular projectile forms are not printed. Not in the formula sheet: learn it.

The horizontal velocity is constant because horizontal acceleration is zero. The vertical component can be positive, zero or negative during the same flight.

Worked example 1: form the model from speed and angle

Question. A ball is launched from $(0,20)\,\mathrm m$ at $10\,\mathrm{m\,s^{-1}}$ , at $30^\circ$ above horizontal. Neglect air resistance and use $g=10\,\mathrm{m\,s^{-2}}$ . Determine its initial velocity, velocity at time $t$ , and position at time $t$ , before impact.

Plan. Resolve the launch velocity, then use the fixed upward axis for both vector equations.

\begin{aligned} u_x&=10\cos30^\circ=5\sqrt3,\qquad u_y=10\sin30^\circ=5 && \text{resolve launch}\\ \mathbf v&=5\sqrt3\,\mathbf i+(5-10t)\mathbf j && \text{apply gravity}\\ \mathbf r&=5\sqrt3\,t\mathbf i+(20+5t-5t^2)\mathbf j && \text{include starting height} \end{aligned}

Why this step? The starting height belongs in position, while the launch angle determines velocity components.

Answer. $\mathbf u=5\sqrt3\,\mathbf i+5\mathbf j\,\mathrm{m\,s^{-1}}$ , with $\mathbf v$ and $\mathbf r$ as displayed, in SI units during free flight.

Check. At $t=0$ , the equations recover $\mathbf r=20\mathbf j$ and the resolved launch velocity; differentiating position gives velocity.

Change only the angle to thirty degrees below horizontal. The vertical launch component changes sign; gravity does not.

Worked example 2: a downward launch

Question. A ball is launched from $(0,20)\,\mathrm m$ at $10\,\mathrm{m\,s^{-1}}$ , at $30^\circ$ below horizontal. Neglect air resistance and use $g=10\,\mathrm{m\,s^{-2}}$ . Determine its initial velocity, velocity at time $t$ , and position at time $t$ , before impact.

Plan. Resolve with a negative vertical launch component, keeping upwards positive.

\begin{aligned} u_x&=5\sqrt3,\qquad u_y=-5 && \text{downward launch component}\\ \mathbf v&=5\sqrt3\,\mathbf i+(-5-10t)\mathbf j && \text{gravity remains downward}\\ \mathbf r&=5\sqrt3\,t\mathbf i+(20-5t-5t^2)\mathbf j && \text{new vertical motion} \end{aligned}

Why this step? Changing launch direction cannot reverse the direction of gravity.

Answer. $\mathbf u=5\sqrt3\,\mathbf i-5\mathbf j\,\mathrm{m\,s^{-1}}$ ; the displayed $\mathbf v$ and $\mathbf r$ describe the downward launch before impact, in SI units.

Check. The derivative of $20-5t-5t^2$ is $-5-10t$ , which is negative for all $t\ge0$ , as a downward launch requires.

For a horizontal launch $u_y=0$ . From height $5\,\mathrm m$ , using $g=10$ , its ground equation is $5-5t^2=0$ , giving a one-second fall independently of horizontal speed. That independence follows from the assumptions, rather than from a cancellation between the horizontal and vertical motions.

Watch out

Wrong: use $g=9.8$ automatically when a question gives another value. Right: use the stated acceleration and the sign dictated by your axes.

Change only the angle again and form the full model.

Position, speed and direction during flight

At a specified horizontal position, find the time first, then use that same time in both position and velocity components.

Consider flight under a stated $g=4\,\mathrm{m\,s^{-2}}$ , with initial velocity $6\mathbf i+12\mathbf j\,\mathrm{m\,s^{-1}}$ from level ground. The model gives $x=6t$ , $y=12t-2t^2$ and $\mathbf v=6\mathbf i+(12-4t)\mathbf j$ . Its return to ground is at $t=6$ .

Speed and flight direction

For velocity $\mathbf v=v_x\mathbf i+v_y\mathbf j$ ,

|\mathbf v|=\sqrt{v_x^2+v_y^2}

For $v_x>0$ , an angle to horizontal satisfies $\tan\alpha=|v_y|/v_x$ ; use “above” or “below” according to the sign of $v_y$ . Not in the formula sheet: learn it.

A vertical velocity component of zero means horizontal motion if $v_x\ne0$ . Position gives a different direction: $y/x$ is the slope from the launch point, not the flight direction.

Worked example 3: reaching a horizontal coordinate

Question. A projectile starts at the origin with velocity $6\mathbf i+12\mathbf j\,\mathrm{m\,s^{-1}}$ . Neglect air resistance and use $g=4\,\mathrm{m\,s^{-2}}$ . At horizontal coordinate $x=12\,\mathrm m$ , determine its position, velocity, exact speed and direction. Give the direction angle to one decimal place.

Plan. Recover time from horizontal motion and evaluate the vector equations.

\begin{aligned} t&=12/6=2 && \text{horizontal motion}\\ \mathbf r&=12\mathbf i+[12(2)-2(2^2)]\mathbf j=12\mathbf i+16\mathbf j && \text{position}\\ \mathbf v&=6\mathbf i+4\mathbf j && \text{velocity at the same time}\\ |\mathbf v|&=\sqrt{36+16}=2\sqrt{13} && \text{speed}\\ \alpha&=\tan^{-1}(4/6)=33.7^\circ && \text{upward component} \end{aligned}

Why this step? A horizontal coordinate determines time here because horizontal velocity is constant and non-zero.

Answer. Position is $12\mathbf i+16\mathbf j\,\mathrm m$ , velocity is $6\mathbf i+4\mathbf j\,\mathrm{m\,s^{-1}}$ , speed is $2\sqrt{13}\,\mathrm{m\,s^{-1}}$ , and direction is $33.7^\circ$ above horizontal.

Check. The time $2$ lies inside $0\le t\le6$ , before the highest point at $t=3$ .

Change only the specified horizontal coordinate to twenty-four metres. The projectile reaches the same height on its descent.

Worked example 4: the descending branch

Question. For the same launch $6\mathbf i+12\mathbf j\,\mathrm{m\,s^{-1}}$ , no air resistance and $g=4\,\mathrm{m\,s^{-2}}$ , determine position, velocity, exact speed and direction at $x=24\,\mathrm m$ . Give the direction angle to one decimal place.

Plan. Use the changed horizontal coordinate to obtain a new common time.

\begin{aligned} t&=24/6=4 && \text{new horizontal position}\\ \mathbf r&=24\mathbf i+[48-32]\mathbf j=24\mathbf i+16\mathbf j && \text{position}\\ \mathbf v&=6\mathbf i-4\mathbf j && \text{negative vertical component}\\ |\mathbf v|&=2\sqrt{13},\qquad \alpha=\tan^{-1}(4/6)=33.7^\circ && \text{speed and angle} \end{aligned}

Why this step? Equal heights can occur at two different times, with opposite vertical velocities.

Answer. Position is $24\mathbf i+16\mathbf j\,\mathrm m$ , velocity is $6\mathbf i-4\mathbf j\,\mathrm{m\,s^{-1}}$ , speed is $2\sqrt{13}\,\mathrm{m\,s^{-1}}$ , and direction is $33.7^\circ$ below horizontal.

Check. Times two and four are equally spaced about the apex time three; their equal heights and opposite vertical components agree.

A specified height can give two permissible roots. Evaluate the velocity at each unless a rising or falling branch is specified, and reject times after first impact. A ball can have positive height and negative vertical velocity.

Watch out

Wrong: use the angle $\tan^{-1}(y/x)$ as the direction of flight. Right: use the velocity components, with the correct reference direction and vertical sense.

Move the specified horizontal coordinate once more, to the apex.

Flight time, horizontal range and greatest height

Choose the landing level before using a flight-time formula, and distinguish height above ground from rise above launch.

For launch height $y_0$ , ground impact on a horizontal level $y=0$ satisfies $y_0+u_yT-\tfrac12gT^2=0$ . Select the first positive impact root. With $u_x>0$ , horizontal range is $u_xT$ ; it is not distance along the curved flight.

Same-level range and greatest rise

For $u_x>0$ , $u_y>0$ , no earlier obstruction and landing at the launch level,

T=\frac{2u_y}{g},\qquad R=\frac{2u_xu_y}{g}

At an attained apex,

t_H=\frac{u_y}{g},\qquad H=\frac{u_y^2}{2g}

Here $H$ is rise above launch. These are derived from the component equations. Not in the formula sheet: learn it.

For the first result, factor $0=T(u_y-\tfrac12gT)$ , rejecting the release root. For the second pair, solve $u_y-gt_H=0$ and substitute into height: $H=u_y(u_y/g)-\tfrac12g(u_y/g)^2$ . If launch is $y_0$ above ground, greatest height is $y_0+H$ , provided impact has not interrupted the ascent.

For fixed launch speed $U$ , same-level range is $R=U^2\sin(2\theta)/g$ . Thus $45^\circ$ gives greatest range on that horizontal level, since $\sin(2\theta)\le1$ for $0<\theta<90^\circ$ . This derived result is not printed on the sheet and is not a shortcut for sloping ground.

[DIAGRAM: asset_name: A projectile with horizontal velocity at its apex; asset_slug: projectile_flight; description: Equal-scale trajectory x=12t, y=16t-5t squared from launch to same-level landing at x=38.4 metres, with apex (19.2,12.8), upward-right launch tangent and horizontal apex velocity; caption: A projectile with horizontal velocity at its apex]

For this figure alone, $\mathbf u=12\mathbf i+16\mathbf j$ and $g=10$ , giving flight time $3.2$ , range $38.4$ , apex time $1.6$ and greatest rise $12.8$ , in SI units. The horizontal velocity survives at the top.

Worked example 5: level-ground flight

Question. A projectile starts from horizontal ground with velocity $6\mathbf i+12\mathbf j\,\mathrm{m\,s^{-1}}$ . Neglect air resistance and use $g=4\,\mathrm{m\,s^{-2}}$ . Determine its flight time, horizontal range and greatest height above ground.

Plan. Solve the actual ground equation, then find the apex from vertical velocity.

\begin{aligned} 0&=12T-2T^2=2T(6-T) && \text{ground impact}\\ T&=6,\qquad R=6(6)=36 && \text{post-launch root}\\ t_H&=12/4=3 && \text{vertical velocity zero}\\ y(3)&=36-18=18 && \text{greatest height} \end{aligned}

Why this step? The two roots represent release and return; only the second is the flight duration.

Answer. Flight time is $6\,\mathrm s$ , horizontal range is $36\,\mathrm m$ , and greatest height is $18\,\mathrm m$ .

Check. The apex lies halfway through this same-level flight.

Change only the launch height to fourteen metres above ground. The same three requests now need an unequal-level calculation.

Worked example 6: a lower landing level

Question. A projectile starts $14\,\mathrm m$ above horizontal ground with velocity $6\mathbf i+12\mathbf j\,\mathrm{m\,s^{-1}}$ . Neglect air resistance and use $g=4\,\mathrm{m\,s^{-2}}$ . Determine its flight time, horizontal range and greatest height above ground.

Plan. Include fourteen metres in the height equation, and keep the unchanged apex time.

\begin{aligned} 0&=14+12T-2T^2 && \text{ground is below launch}\\ 0&=(T-7)(T+1) && \text{divide and factor}\\ T&=7,\qquad R=6(7)=42 && \text{positive impact time}\\ y(3)&=14+36-18=32 && \text{apex above ground} \end{aligned}

Why this step? The rise above launch is unchanged, but the height above ground and the total fall time have increased.

Answer. Flight time is $7\,\mathrm s$ , horizontal range is $42\,\mathrm m$ , and greatest height above ground is $32\,\mathrm m$ .

Check. The other root $-1$ is before launch, and the apex time three occurs before impact.

The free-flight apex need not be attained if an elevated surface intercepts the projectile first. The final part uses that observation to resolve the opening question.

Watch out

Wrong: add launch height to the range or use $2u_y/g$ for a lower landing level. Right: launch height enters the vertical equation; range then follows from the resulting time.

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My1–My5 - Projectile motion and trajectories

Assumptions and component equations

Position, speed and direction during flight

Flight time, horizontal range and greatest height

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