Q5 - Projectile motion | AQA A-Level Mathematics

Model a projectile in a vertical plane using horizontal and vertical components, derive its trajectory, and interpret heights, ranges and impact times. You will use the assumptions of the model explicitly and choose the physical solution when an equation gives several possible times.

The projectile component model

Could launches at $30^\circ$ and $60^\circ$ , with the same speed, land at the same horizontal distance? They can in an ideal equal-level flight, even though one flies much higher. The explanation begins with two motions sharing one clock.

You already know

Resolve a velocity of magnitude $U$ at angle $\theta$ above the horizontal into $U\cos\theta$ horizontally and $U\sin\theta$ vertically. Constant-acceleration displacement and velocity equations apply componentwise, using the same elapsed time.

Treat the object as a particle, ignoring its size and spin. Separately, neglect air resistance and wind, and assume constant downward gravitational acceleration. The particle assumption does not imply the absence of air resistance. The model describes a fixed vertical plane from launch until the first collision with the ground or an obstacle.

Choose $\mathbf i$ horizontally forward and $\mathbf j$ vertically upwards, with launch at $(0,h)$ . Then $\mathbf r_0=h\mathbf j$ , $\mathbf u=U\cos\theta\,\mathbf i+U\sin\theta\,\mathbf j$ and $\mathbf a=-g\mathbf j$ . The vector constant-acceleration equations are supplied in the formula booklet; the projectile specialisations below are derived from them.

Projectile component equations

Before the first collision,

\begin{aligned} x&=U\cos\theta\,t \\ y&=h+U\sin\theta\,t-\tfrac12gt^2 \\ v_x&=U\cos\theta \\ v_y&=U\sin\theta-gt \end{aligned}

Here $t$ is time after launch, with positive vertical direction upwards. Not in the formula booklet: learn it.

Horizontal velocity stays constant while vertical velocity changes. At an upward flight's highest point, only the vertical component is zero: unless launch is purely vertical, the object is still moving horizontally. For a launch at angle $\alpha$ from the upward vertical, the initial components are $U\sin\alpha$ horizontally and $U\cos\alpha$ vertically. Resolution, vector magnitude and direction ratios are recall results, not supplied booklet formulae; draw the right triangle when the angle's reference line changes.

Worked example 1: a level-ground launch

Question. A ball is launched from level ground at $20\,\mathrm{m\,s^{-1}}$ , at $30^\circ$ above the horizontal. Use $g=9.8\,\mathrm{m\,s^{-2}}$ , the particle model and no air resistance. Form its position and velocity functions, then describe it at $t=1$ .

Plan. Resolve initial velocity, apply the component equations, then substitute the common time.

\begin{aligned} u_x&=20\cos30^\circ=10\sqrt3,&u_y&=20\sin30^\circ=10 \\ x&=10\sqrt3\,t,&y&=10t-4.9t^2 && \text{position components} \\ \mathbf v&=10\sqrt3\,\mathbf i+(10-9.8t)\mathbf j && \text{velocity components} \\ \mathbf r(1)&=10\sqrt3\,\mathbf i+5.1\mathbf j && \text{same time in both components} \\ v_y(1)&=0.2>0 && \text{still rising} \end{aligned}

Why resolve first? Gravity acts vertically, so it changes the vertical velocity rather than subtracting $gt$ from the initial speed.

Answer. The vector functions are $\mathbf r=10\sqrt3\,t\mathbf i+(10t-4.9t^2)\mathbf j$ metres and $\mathbf v=10\sqrt3\,\mathbf i+(10-9.8t)\mathbf j\,\mathrm{m\,s^{-1}}$ . At $1$ second the ball is rising at height $5.1\,\mathrm m$ .

Check. At zero, position is the origin and the initial velocity components give magnitude $\sqrt{300+100}=20\,\mathrm{m\,s^{-1}}$ .

Worked example 2 changes only the launch height. Velocity is unchanged at a common time because the initial velocity and acceleration are unchanged.

Worked example 2: launch above the ground

Question. The same ball is launched from $2.5\,\mathrm m$ above horizontal ground, with the same initial velocity and assumptions. Find its position and velocity at $t=1$ .

Plan. Add the launch height to vertical position, while keeping both velocity components unchanged.

\begin{aligned} x(1)&=10\sqrt3 && \text{same horizontal motion} \\ y(1)&=2.5+10-4.9=7.6 && \text{include launch height} \\ \mathbf v(1)&=10\sqrt3\,\mathbf i+0.2\mathbf j && \text{unchanged velocity at }1 \end{aligned}

Why does height affect position but not velocity? It adds a constant vertical offset, whose time derivative is zero.

Answer. Position is $10\sqrt3\,\mathbf i+7.6\mathbf j$ metres and velocity is $10\sqrt3\,\mathbf i+0.2\mathbf j\,\mathrm{m\,s^{-1}}$ .

Check. The height is exactly $2.5\,\mathrm m$ above the first trajectory at the same time; it remains before ground contact.

Use the value of $g$ directed in a question, such as $9.8$ , $9.81$ or $10\,\mathrm{m\,s^{-2}}$ . Retain unrounded intermediate values, then follow any requested precision; AQA's projectile guidance otherwise uses the accuracy of the stated $g$ .

Watch out

Wrong: “The velocity is zero at the top.” Only $v_y=0$ ; $v_x=U\cos\theta$ remains. Right: for $0<\theta<90^\circ$ , velocity at the highest point is a nonzero horizontal vector.

Now build a fresh model with a different launch velocity. Keep the initial height in position, and leave exact surds in the functions.

Maximum height, flight time and range

Use vertical motion to find the event time, then use that same time in horizontal motion.

For an upward launch, $v_y=0$ gives $t_{\max}=U\sin\theta/g$ . Substituting this into vertical position gives height above ground $h+U^2\sin^2\theta/(2g)$ . The term $U^2\sin^2\theta/(2g)$ is the rise above launch, not automatically the height above ground.

If landing is at the launch level, put $y=h$ . Then $0=t(U\sin\theta-gt/2)$ . The root $t=0$ is launch; the later root gives the flight time.

Equal-level flight

For $0<\theta<90^\circ$ , with landing at launch level,

T=\frac{2U\sin\theta}{g} \qquad R=U\cos\theta\,T=\frac{U^2\sin2\theta}{g}

The rise to maximum height is

H=\frac{U^2\sin^2\theta}{2g}

These use the ideal particle model with no air resistance and constant $g$ . Not in the formula booklet: learn it.

The double-angle identity $\sin2\theta=2\sin\theta\cos\theta$ is also a recall result. For fixed $U,g$ , equal-level range is greatest at $45^\circ$ , since $\sin2\theta\le1$ . Complementary launch angles have equal ranges because $\sin(180^\circ-2\theta)=\sin2\theta$ . Neither conclusion applies automatically when landing is lower than launch.

Worked example 3: height, time and range

Question. For the $20\,\mathrm{m\,s^{-1}}$ , $30^\circ$ level-ground launch, find time to maximum height, maximum height, flight time and horizontal range. Use $g=9.8\,\mathrm{m\,s^{-2}}$ and give final approximations to two significant figures.

Plan. Use $v_y=0$ for the top, $y=0$ for later landing, then use horizontal velocity for range.

\begin{aligned} t_{\max}&=\frac{10}{9.8}=\frac{50}{49} && \text{zero vertical velocity} \\ H&=\frac{10^2}{2(9.8)}=\frac{250}{49} && \text{vertical rise} \\ T&=\frac{2(10)}{9.8}=\frac{100}{49} && \text{later ground root} \\ R&=10\sqrt3\left(\frac{100}{49}\right)=\frac{1000\sqrt3}{49} && \text{horizontal speed times time} \end{aligned}

Why retain these fractions? Rounding $T$ before multiplying would lose accuracy in the range.

Answer. The final values are $1.0\,\mathrm s$ , $5.1\,\mathrm m$ , $2.0\,\mathrm s$ and $35\,\mathrm m$ , respectively.

Check. Equal-level flight has $T=2t_{\max}$ , and the range formula gives the same exact $1000\sqrt3/49$ metres.

Worked example 4 changes only the angle to its complement. This tests the opening puzzle without changing launch speed or gravitational acceleration.

Worked example 4: a higher flight with the same range

Question. Repeat the calculation for a $60^\circ$ launch at $20\,\mathrm{m\,s^{-1}}$ , landing at its launch level, with $g=9.8\,\mathrm{m\,s^{-2}}$ . Give final approximations to two significant figures.

Plan. Use the new vertical component $10\sqrt3$ and horizontal component $10$ .

\begin{aligned} t_{\max}&=\frac{10\sqrt3}{9.8}=\frac{50\sqrt3}{49} && \text{vertical velocity zero} \\ H&=\frac{(10\sqrt3)^2}{19.6}=\frac{750}{49} && \text{vertical rise} \\ T&=\frac{100\sqrt3}{49} && \text{twice ascent time} \\ R&=10\left(\frac{100\sqrt3}{49}\right)=\frac{1000\sqrt3}{49} && \text{horizontal distance} \end{aligned}

Why is the range unchanged? The longer flight time exactly compensates for the smaller horizontal velocity in this equal-level model.

Answer. Time to the top is $1.8\,\mathrm s$ , height is $15\,\mathrm m$ , flight time is $3.5\,\mathrm s$ and range is $35\,\mathrm m$ .

Check. The exact maximum height is three times that of the $30^\circ$ launch, although the exact ranges agree.

If the ground is below launch, solve $y=0$ using the full height expression. In particular, a horizontal launch above ground can have a positive flight time even though its initial vertical velocity is zero.

Watch out

Wrong: use $T=2U\sin\theta/g$ for every landing height. This formula finds a return to launch level. Right: form the height equation for the actual landing level, then choose the physically relevant root.

Now use the angle that maximises equal-level range. Calculate directly from components to check the result.

Trajectory equations and obstacles

Eliminating time gives height directly from horizontal position, which is useful for obstacles and for recovering a launch from known points on its path.

For nonzero horizontal velocity, $t=x/(U\cos\theta)$ . Substitute this into vertical position to obtain the trajectory. A purely vertical launch instead has $x=0$ throughout; division by $\cos90^\circ$ would be invalid.

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Q5 - Projectile motion

The projectile component model

Maximum height, flight time and range

Trajectory equations and obstacles

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