15.6 Motion under gravity and projectiles using vectors — Practise with help

Work each one first, closed book. Resolve the initial velocity before you touch any formula, and write down the positive direction. The two components share one time $t$ .

Warm-up: remember

Resolve a velocity of $10$ m/s at $30^\circ$ to the horizontal into its components.
Write $20\cos 45^\circ$ as an exact surd.
Solve $0 = 10 - 9.8t$ , giving $t$ to three significant figures.

Answers

Horizontal $= 10\cos 30^\circ = 5\sqrt{3} \approx 8.66$ m/s, and vertical $= 10\sin 30^\circ = 5$ m/s.
$20 \times \dfrac{\sqrt{2}}{2} = 10\sqrt{2}$ .
$t = \dfrac{10}{9.8} \approx 1.02$ .

If question 1 was hard, revise resolving vectors from Section 5.12 before you continue. Every projectile on this page starts with that step.

Practise with help

1. A ball is projected at $24$ m/s at $30^\circ$ to the horizontal. Find the horizontal and vertical components of its initial velocity.

Hint

Use $u_x = U\cos\theta$ and $u_y = U\sin\theta$ . The angle is measured from the horizontal.

Next step

Work out $24\cos 30^\circ$ and $24\sin 30^\circ$ . Keep the surds exact.

Full working

u_x = 24\cos 30^\circ = 24 \times \frac{\sqrt{3}}{2} = 12\sqrt{3} \approx 20.8 \text{ m/s},

u_y = 24\sin 30^\circ = 24 \times \frac{1}{2} = 12 \text{ m/s}.

2. For the ball in question 1, find the time to the greatest height. Take $g = 9.8$ m/s $^2$ .

Hint

At the greatest height the vertical velocity is zero. Use the vertical component only.

Next step

Set $0 = u_y - gt$ with $u_y = 12$ and $g = 9.8$ .

Full working

0 = 12 - 9.8t, \qquad t = \frac{12}{9.8} = \frac{120}{98} = \frac{60}{49} \approx 1.22 \text{ s}.

3. For the ball in question 1, find the total time of flight and the range.

Hint

The ball lands at the height it started from, so the total time is twice the time to the greatest height. The range is the horizontal velocity times the total time.

Next step

Double the time from question 2. Then multiply $u_x = 12\sqrt{3}$ by that total time.

Full working

T = 2 \times \frac{60}{49} = \frac{120}{49} \approx 2.45 \text{ s},

R = u_x T = 12\sqrt{3} \times \frac{120}{49} = \frac{1440\sqrt{3}}{49} \approx 50.9 \text{ m}.

4. For the ball in question 1, find the greatest height.

Hint

Use the vertical component with $v_y = 0$ at the top, and the formula $v_y^2 = u_y^2 - 2gy$ .

Next step

$0 = 12^2 - 2 \times 9.8 \times H$ . Rearrange for $H$ .

Full working

0 = 144 - 19.6H, \qquad H = \frac{144}{19.6} = \frac{360}{49} \approx 7.35 \text{ m}.

5. A ball is kicked at $30$ m/s at $60^\circ$ to the horizontal. Find its vertical velocity and its height after $2$ s. Take $g = 9.8$ m/s $^2$ and upwards as positive.

Hint

Resolve to get $u_y = 30\sin 60^\circ = 15\sqrt{3}$ . Then use $v_y = u_y - gt$ and $y = u_y t - \frac{1}{2}gt^2$ .

Next step

At $t = 2$ : $v_y = 15\sqrt{3} - 19.6$ and $y = 30\sqrt{3} - 19.6$ .

Full working

v_y = 15\sqrt{3} - 9.8 \times 2 = 15\sqrt{3} - 19.6 \approx 6.38 \text{ m/s},

y = 15\sqrt{3} \times 2 - \frac{1}{2} \times 9.8 \times 2^2 = 30\sqrt{3} - 19.6 \approx 32.4 \text{ m}.

The vertical velocity is still positive, so the ball is rising at $2$ s.

6. A ball is hit horizontally at $12$ m/s from the top of a cliff that is $25$ m high. Find the time it takes to reach the ground and the horizontal distance travelled. Take $g = 9.8$ m/s $^2$ .

Hint

The initial vertical velocity is $0$ . Take downwards as positive for the fall and use $s = \frac{1}{2}gt^2$ with $s = 25$ .

Next step

$25 = 4.9t^2$ , so $t^2 = \dfrac{25}{4.9} = \dfrac{250}{49}$ . Then use $x = 12t$ .

Full working

t = \sqrt{\frac{250}{49}} = \frac{5\sqrt{10}}{7} \approx 2.26 \text{ s},

x = 12 \times \frac{5\sqrt{10}}{7} = \frac{60\sqrt{10}}{7} \approx 27.1 \text{ m}.

7. A ball is kicked from the ground at $20$ m/s at $45^\circ$ . Find its height after it has travelled $30$ m horizontally. Take $g = 9.8$ m/s $^2$ .

Hint

First find the time from the horizontal equation $x = u_x t$ , with $u_x = 10\sqrt{2}$ . Then substitute that time into the vertical equation.

Next step

$30 = 10\sqrt{2}\,t$ , so $t = \dfrac{3}{\sqrt{2}} = \dfrac{3\sqrt{2}}{2}$ . Substitute this into $y = 10\sqrt{2}\,t - 4.9t^2$ .

Full working

t = \frac{30}{10\sqrt{2}} = \frac{3\sqrt{2}}{2},

y = 10\sqrt{2} \times \frac{3\sqrt{2}}{2} - 4.9 \times \left(\frac{3\sqrt{2}}{2}\right)^2 = 30 - 4.9 \times \frac{18}{4} = 30 - 22.05 = 7.95 \text{ m}.

The ball is about $7.95$ m high when it has travelled $30$ m horizontally.