Projectiles – A Level Maths Year 2

1 The Projectile Model

A projectile is a particle moving freely under gravity alone — a kicked ball, a stone thrown from a cliff, a jet of water. DfE content area Q · Kinematics asks you to model this motion in two dimensions.

The key idea: treat the horizontal and vertical motions separately. They share only one thing — the time $t$.
• Horizontally there is no force (no air resistance), so the velocity is constant.
• Vertically there is constant acceleration $g$ downwards, so the Year 1 suvat equations apply.

Projected with speed $u$ at angle $\theta$ above the horizontal

Initial components: horizontal $u\cos\theta$, vertical $u\sin\theta$
$x = (u\cos\theta)\,t$ $y = (u\sin\theta)\,t - \tfrac12 g t^2$
$\dot{x} = u\cos\theta$ $\dot{y} = u\sin\theta - gt$

Choose a sign convention and stick to it. Taking upwards as positive means the vertical acceleration is $-9.8$ m s$^{-2}$ and a point below the launch height has a negative $y$.

2 Projected Horizontally

When $\theta = 0$ the initial vertical velocity is zero, so the vertical motion is just a dropped object.

Worked Example 1 — Off a cliff

A stone is thrown horizontally at $15$ m s$^{-1}$ from the top of a cliff $40$ m above the sea. Find the time to reach the sea, the horizontal distance travelled, and the speed and direction of motion as it hits the water.

①Vertical (down positive): $s = 40$, $u = 0$, $a = 9.8$. $40 = 4.9t^2 \Rightarrow t = 2.857$ s, so $2.86$ s.

②Horizontal: $x = 15 \times 2.857 = 42.9$ m.

③Vertical velocity at impact: $v_y = 9.8 \times 2.857 = 28.0$ m s$^{-1}$ downwards; horizontal still $15$ m s$^{-1}$.

④Speed $\sqrt{15^2 + 28^2} = 31.8$ m s$^{-1}$, at $\tan^{-1}\!\left(\tfrac{28}{15}\right) = 61.8^\circ$ below the horizontal.

3 Time of Flight, Range and Maximum Height

For a projectile launched from and landing on level ground:

Quantity	Condition	Result
Time to the top	vertical velocity $= 0$	$t = \dfrac{u\sin\theta}{g}$
Maximum height $H$	$0 = (u\sin\theta)^2 - 2gH$	$H = \dfrac{u^2\sin^2\theta}{2g}$
Time of flight $T$	$y = 0$, $t \neq 0$	$T = \dfrac{2u\sin\theta}{g}$
Range $R$	$x$ at time $T$	$R = \dfrac{u^2\sin 2\theta}{g}$

Deriving the range: $R = u\cos\theta \times \dfrac{2u\sin\theta}{g} = \dfrac{u^2(2\sin\theta\cos\theta)}{g} = \dfrac{u^2\sin 2\theta}{g}$, using the double-angle identity. Since $\sin 2\theta \leqslant 1$, the maximum range is $\dfrac{u^2}{g}$, at $\theta = 45^\circ$.

These formulae only work on level ground, and exam questions often ask you to derive them. It is safer to work from the component equations every time than to memorise results.

Worked Example 2 — A ball kicked from the ground

A ball is kicked from level ground at $20$ m s$^{-1}$ at $35^\circ$ above the horizontal. Find the time of flight, the range and the greatest height.

①Components: $u_x = 20\cos 35^\circ = 16.38$, $u_y = 20\sin 35^\circ = 11.47$ m s$^{-1}$.

②Lands when $y = 0$: $11.47t - 4.9t^2 = 0 \Rightarrow t = \dfrac{11.47}{4.9} = 2.34$ s.

③Range $= 16.38 \times 2.341 = 38.4$ m.

④Top: $0 = 11.47^2 - 2(9.8)H \Rightarrow H = 6.71$ m.

Worked Example 3 — Velocity at a given time

For the ball in Worked Example 2, find its speed and direction of motion $1.5$ s after it is kicked.

①$v_x = 16.38$ m s$^{-1}$ (unchanged).

②$v_y = 11.47 - 9.8 \times 1.5 = -3.23$ m s$^{-1}$ — negative, so it is already coming down.

③Speed $\sqrt{16.38^2 + 3.23^2} = 16.7$ m s$^{-1}$.

④Direction $\tan^{-1}\!\left(\tfrac{3.23}{16.38}\right) = 11.1^\circ$ below the horizontal.

Worked Example 4 — Finding the angle from the range

A particle projected at $30$ m s$^{-1}$ from level ground lands $60$ m away. Find the possible angles of projection.

①$60 = \dfrac{30^2\sin 2\theta}{9.8} \Rightarrow \sin 2\theta = \dfrac{588}{900} = 0.6533$

②$2\theta = 40.79^\circ$ or $180^\circ - 40.79^\circ = 139.21^\circ$

③$\theta = 20.4^\circ$ or $69.6^\circ$.

The two angles are complementary (they add to $90^\circ$). A low, flat shot and a high lob reach the same point.

4 Projection from a Height, and Clearing Obstacles

If the launch and landing heights differ, the level-ground formulae fail. Go back to $y = (u\sin\theta)t - \tfrac12gt^2$ and solve a quadratic in $t$.

Worked Example 5 — Thrown upwards from a cliff

A stone is thrown at $15$ m s$^{-1}$ at $30^\circ$ above the horizontal from the top of a cliff $25$ m above the sea. Find how long it takes to hit the sea and how far from the foot of the cliff it lands.

①Up positive, origin at the launch point. The sea is at $y = -25$.

②$-25 = 15\sin 30^\circ\,t - 4.9t^2 \Rightarrow 4.9t^2 - 7.5t - 25 = 0$

③$t = \dfrac{7.5 + \sqrt{7.5^2 + 4(4.9)(25)}}{9.8} = \dfrac{7.5 + 23.37}{9.8} = 3.15$ s (the negative root is rejected).

④$x = 15\cos 30^\circ \times 3.150 = 40.9$ m.

Worked Example 6 — Clearing a wall

A ball is projected from ground level at $25$ m s$^{-1}$ at $40^\circ$ to the horizontal towards a wall $6$ m high, $30$ m away. Does it clear the wall?

①Use the horizontal motion to find when it reaches the wall: $t = \dfrac{30}{25\cos 40^\circ} = 1.567$ s.

②Height then: $y = 25\sin 40^\circ(1.567) - 4.9(1.567)^2 = 25.18 - 12.03 = 13.1$ m.

③$13.1 > 6$, so it clears the wall, by about $7.1$ m.

5 The Equation of the Trajectory

Eliminate $t$ between the $x$ and $y$ equations to get the path as a Cartesian equation. From $x = (u\cos\theta)t$, $t = \dfrac{x}{u\cos\theta}$. Substitute:

Trajectory of a projectile (origin at the point of projection)

$y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$ $=\; x\tan\theta - \dfrac{gx^2}{2u^2}\left(1 + \tan^2\theta\right)$

It is a quadratic in $x$ with a negative $x^2$ coefficient — the path is a parabola. The second form uses $\sec^2\theta = 1 + \tan^2\theta$ and turns "find the angle" problems into a quadratic in $\tan\theta$.

Worked Example 7 — Hitting a target

A particle is projected from $O$ at $20$ m s$^{-1}$. It must pass through the point $(30, 5)$ m. Find the possible angles of projection.

①Let $T = \tan\theta$: $5 = 30T - \dfrac{9.8 \times 900}{2 \times 400}(1 + T^2) = 30T - 11.025(1 + T^2)$

②$11.025T^2 - 30T + 16.025 = 0$

③$T = \dfrac{30 \pm \sqrt{900 - 4(11.025)(16.025)}}{2(11.025)} = \dfrac{30 \pm 13.90}{22.05}$

④$T = 1.991$ or $0.7301$, so $\theta = 63.3^\circ$ or $36.1^\circ$.

If the discriminant were negative, the target would be out of reach at that speed.

6 Quick Reference

Components

$u\cos\theta$ across, $u\sin\theta$ up.

Horizontal

Constant velocity: $x = (u\cos\theta)t$.

Vertical

suvat with $a = -g$.

Top

Vertical velocity $= 0$.

Level ground

$T = \dfrac{2u\sin\theta}{g}$, $R = \dfrac{u^2\sin2\theta}{g}$.

Max height

$H = \dfrac{u^2\sin^2\theta}{2g}$.

Max range

At $45^\circ$: $u^2/g$.

From a height

Solve $y = -h$ as a quadratic in $t$.

Trajectory

$y = x\tan\theta - \dfrac{gx^2}{2u^2}(1+\tan^2\theta)$.

Speed

$\sqrt{v_x^2 + v_y^2}$, direction $\tan^{-1}(v_y/v_x)$.

7 Practice Questions

Use $g = 9.8$ m s$^{-2}$, ignore air resistance, and give answers to 3 significant figures where they are not exact.

Question 1

A stone is thrown horizontally at $8$ m s$^{-1}$ from a point $20$ m above level ground. Find the time it takes to land and the horizontal distance travelled.

▶ Show solution

$20 = 4.9t^2 \Rightarrow t = \sqrt{\dfrac{20}{4.9}} = 2.02$ s.

$x = 8 \times 2.020 = 16.2$ m.

Question 2

A particle is projected from level ground at $30$ m s$^{-1}$ at $50^\circ$ above the horizontal. Find the time of flight, the range and the greatest height.

▶ Show solution

$u_x = 30\cos 50^\circ = 19.28$, $u_y = 30\sin 50^\circ = 22.98$.

Time of flight: $T = \dfrac{2(22.98)}{9.8} = 4.69$ s.

Range: $19.28 \times 4.690 = 90.4$ m.

Greatest height: $H = \dfrac{22.98^2}{2 \times 9.8} = 26.9$ m.

Question 3

For the particle in Question 2, find its speed and direction of motion $2$ s after projection.

▶ Show solution

$v_x = 19.28$; $v_y = 22.98 - 19.6 = 3.38$ m s$^{-1}$ (still rising).

Speed $= \sqrt{19.28^2 + 3.38^2} = 19.6$ m s$^{-1}$.

Direction $\tan^{-1}\!\left(\tfrac{3.38}{19.28}\right) = 9.95^\circ$ above the horizontal.

Question 4

Show that a particle projected from level ground with speed $u$ at angle $\theta$ has range $\dfrac{u^2\sin 2\theta}{g}$, and deduce the angle giving the greatest range.

▶ Show solution

Vertically: $y = u\sin\theta\,t - \tfrac12gt^2 = t\left(u\sin\theta - \tfrac12gt\right)$. $y = 0$ when $t = 0$ or $t = \dfrac{2u\sin\theta}{g}$.

Horizontally: $R = u\cos\theta \times \dfrac{2u\sin\theta}{g} = \dfrac{u^2 \cdot 2\sin\theta\cos\theta}{g} = \dfrac{u^2\sin 2\theta}{g}$.

$\sin 2\theta$ has maximum $1$ when $2\theta = 90^\circ$, so the greatest range is at $\theta = 45^\circ$.

Question 5

A ball is kicked from level ground at $18$ m s$^{-1}$ at angle $\alpha$ above the horizontal, where $\tan\alpha = \tfrac34$. For how long is the ball more than $4$ m above the ground?

▶ Show solution

$\sin\alpha = 0.6$, so $u_y = 18 \times 0.6 = 10.8$ m s$^{-1}$.

$4 = 10.8t - 4.9t^2 \Rightarrow 4.9t^2 - 10.8t + 4 = 0$.

$t = \dfrac{10.8 \pm \sqrt{116.64 - 78.4}}{9.8} = \dfrac{10.8 \pm 6.184}{9.8}$, so $t = 0.471$ s or $1.733$ s.

Time above $4$ m $= 1.733 - 0.471 = 1.26$ s.

Question 6

A ball is thrown from the top of a building $30$ m high at $12$ m s$^{-1}$ at $20^\circ$ above the horizontal. Find the time to reach the ground, the horizontal distance travelled and the speed on landing.

▶ Show solution

$u_x = 12\cos 20^\circ = 11.28$, $u_y = 12\sin 20^\circ = 4.104$.

$-30 = 4.104t - 4.9t^2 \Rightarrow 4.9t^2 - 4.104t - 30 = 0$.

$t = \dfrac{4.104 + \sqrt{4.104^2 + 588}}{9.8} = \dfrac{4.104 + 24.59}{9.8} = 2.93$ s.

Distance: $11.28 \times 2.928 = 33.0$ m.

$v_y = 4.104 - 9.8(2.928) = -24.59$; speed $= \sqrt{11.28^2 + 24.59^2} = 27.1$ m s$^{-1}$.

(Check: $v^2 = u^2 + 2gh = 144 + 588 = 732$, $v = 27.1$ ✓ — the speed does not depend on the angle.)

Question 7

A particle projected from $O$ at $14$ m s$^{-1}$ passes through the point $(10, 2.5)$ m. Show that $\tan^2\theta - 4\tan\theta + 2 = 0$ and hence find the two possible angles of projection.

▶ Show solution

$\dfrac{gx^2}{2u^2} = \dfrac{9.8 \times 100}{2 \times 196} = 2.5$.

$2.5 = 10\tan\theta - 2.5(1 + \tan^2\theta) \Rightarrow 2.5\tan^2\theta - 10\tan\theta + 5 = 0$.

Divide by $2.5$: $\tan^2\theta - 4\tan\theta + 2 = 0$ ✓

$\tan\theta = 2 \pm \sqrt{2}$, i.e. $3.414$ or $0.5858$.

$\theta = 73.7^\circ$ or $30.4^\circ$.

Question 8

A golf ball is struck from level ground at $35$ m s$^{-1}$ at $25^\circ$ to the horizontal. A tree $12$ m high stands $50$ m away. Does the ball clear the tree?

▶ Show solution

Time to reach the tree: $t = \dfrac{50}{35\cos 25^\circ} = 1.576$ s.

Height: $y = 35\sin 25^\circ(1.576) - 4.9(1.576)^2 = 23.31 - 12.17 = 11.1$ m.

$11.1 < 12$, so the ball does not clear the tree (it hits it about $0.9$ m below the top).

Question 9

A dart is thrown horizontally from a height of $1.8$ m and hits the board $2.4$ m away at a height of $1.6$ m. Find the speed of throw.

▶ Show solution

Vertical drop $= 0.2$ m: $0.2 = 4.9t^2 \Rightarrow t = 0.2020$ s.

Horizontal: $u = \dfrac{2.4}{0.2020} = 11.9$ m s$^{-1}$.

Question 10

A ball is projected from a point $A$, $1.5$ m above level ground, with speed $U$ m s$^{-1}$ at angle $\alpha$ above the horizontal, where $\tan\alpha = \tfrac43$. Its greatest height above the ground is $8$ m.

(a) Find $U$. (b) Find the time from projection until the ball hits the ground. (c) Find the horizontal distance travelled. (d) Find the speed of the ball as it hits the ground. (e) Give one way in which the model could be refined.

▶ Show solution

$\sin\alpha = 0.8$, $\cos\alpha = 0.6$.

(a) The ball rises $8 - 1.5 = 6.5$ m. At the top $v_y = 0$: $0 = (0.8U)^2 - 2(9.8)(6.5)$.

$0.8U = \sqrt{127.4} = 11.29 \Rightarrow U = 14.1$ m s$^{-1}$.

(b) $-1.5 = 11.29t - 4.9t^2 \Rightarrow 4.9t^2 - 11.29t - 1.5 = 0$.

$t = \dfrac{11.29 + \sqrt{127.4 + 29.4}}{9.8} = \dfrac{11.29 + 12.52}{9.8} = 2.43$ s.

(c) $x = 0.6U \times t = 8.465 \times 2.4295 = 20.6$ m.

(d) $v_y = 11.29 - 9.8(2.4295) = -12.52$; speed $= \sqrt{8.465^2 + 12.52^2} = 15.1$ m s$^{-1}$.

(e) Include air resistance; or model the ball's spin; or treat the ball as having size rather than as a particle.

Projectiles · A Level Maths Year 2 · Created with MathJax