15.6 Motion under gravity and projectiles using vectors — Practise

Work through these on your own. Resolve the initial velocity first, choose one positive direction, and keep it. The horizontal and vertical motions share the same time.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A ball is projected at $20$ m/s at $30^\circ$ to the horizontal. Find the horizontal and vertical components of its initial velocity.
A particle is projected at $14$ m/s at $45^\circ$ to the horizontal. Find the horizontal and vertical components of its initial velocity.
A ball is projected at $20$ m/s at $45^\circ$ to the horizontal. Find the time to its greatest height. Take $g = 9.8$ m/s $^2$ .
A ball is projected at $20$ m/s at $45^\circ$ from level ground. Find its range. Take $g = 9.8$ m/s $^2$ .
A ball is projected at $20$ m/s at $45^\circ$ . Find its greatest height. Take $g = 9.8$ m/s $^2$ .
A ball is projected at $24$ m/s at $30^\circ$ from level ground. Find its time of flight. Take $g = 9.8$ m/s $^2$ .
A ball is projected at $24$ m/s at $30^\circ$ from level ground. Find its range. Take $g = 9.8$ m/s $^2$ .
A ball is projected at $24$ m/s at $30^\circ$ . Find its greatest height. Take $g = 9.8$ m/s $^2$ .
A ball is hit horizontally at $15$ m/s from the top of a cliff $20$ m high. Find the time to reach the ground and the horizontal distance travelled. Take $g = 9.8$ m/s $^2$ .
A ball is kicked from the ground at $20$ m/s at $45^\circ$ . Find its height after it has travelled $20$ m horizontally. Take $g = 9.8$ m/s $^2$ .
A ball is projected at $20$ m/s at $45^\circ$ . Find its vertical velocity after $1$ s. Take $g = 9.8$ m/s $^2$ and upwards as positive.
A stone is thrown horizontally at $8$ m/s from a height of $45$ m. Find the time to reach the ground and the horizontal distance travelled. Take $g = 9.8$ m/s $^2$ .

Answers and concise solutions

1. The two components

Answer: Horizontal $= 20\cos 30^\circ = 10\sqrt{3} \approx 17.3$ m/s. Vertical $= 20\sin 30^\circ = 10$ m/s.

2. The two components

Answer: At $45^\circ$ the two components are equal: $14\cos 45^\circ = 14 \times \dfrac{\sqrt{2}}{2} = 7\sqrt{2} \approx 9.90$ m/s, and the same for the vertical component.

3. The time to the greatest height

Answer: The vertical component is $u_y = 20\sin 45^\circ = 10\sqrt{2}$ . At the top $v_y = 0$ , so $t = \dfrac{u_y}{g} = \dfrac{10\sqrt{2}}{9.8} = \dfrac{50\sqrt{2}}{49} \approx 1.44$ s.

4. The range

Answer: Over level ground the range is $R = \dfrac{U^2 \sin 2\theta}{g}$ . Here $2\theta = 90^\circ$ and $\sin 90^\circ = 1$ , so $R = \dfrac{400}{9.8} = \dfrac{2000}{49} \approx 40.8$ m.

5. The greatest height

Answer: The vertical component is $u_y = 10\sqrt{2}$ . The greatest height is $\dfrac{u_y^2}{2g} = \dfrac{200}{19.6} = \dfrac{500}{49} \approx 10.2$ m.

6. The time of flight

Answer: The vertical component is $u_y = 24\sin 30^\circ = 12$ m/s. The time of flight is $\dfrac{2u_y}{g} = \dfrac{24}{9.8} = \dfrac{120}{49} \approx 2.45$ s.

7. The range

Answer: Use $R = \dfrac{U^2 \sin 2\theta}{g}$ with $2\theta = 60^\circ$ : $R = \dfrac{576 \times \frac{\sqrt{3}}{2}}{9.8} = \dfrac{288\sqrt{3}}{9.8} = \dfrac{1440\sqrt{3}}{49} \approx 50.9$ m.

8. The greatest height

Answer: The vertical component is $12$ m/s, so the greatest height is $\dfrac{12^2}{2 \times 9.8} = \dfrac{144}{19.6} = \dfrac{360}{49} \approx 7.35$ m.

9. The time and the horizontal distance

Answer: The initial vertical velocity is $0$ . From $20 = \frac{1}{2} \times 9.8 \times t^2$ , $t = \sqrt{\frac{40}{9.8}} = \frac{10\sqrt{2}}{7} \approx 2.02$ s. The horizontal distance is $15 \times \frac{10\sqrt{2}}{7} = \frac{150\sqrt{2}}{7} \approx 30.3$ m.

10. The height

Answer: From $x = u_x t$ , $20 = 10\sqrt{2}\,t$ , so $t = \sqrt{2}$ . Then $y = 10\sqrt{2} \times \sqrt{2} - 4.9 \times 2 = 20 - 9.8 = 10.2$ m. This is the greatest height.

11. The vertical velocity

Answer: The vertical component is $u_y = 10\sqrt{2} \approx 14.14$ m/s. Then $v_y = u_y - gt = 14.14 - 9.8 = 4.34$ m/s. The velocity is positive, so the ball is still rising.

12. The time and the horizontal distance

Answer: The vertical motion starts from rest: $45 = \frac{1}{2} \times 9.8 \times t^2$ , so $t = \sqrt{\frac{90}{9.8}} = \frac{15\sqrt{2}}{7} \approx 3.03$ s. The horizontal distance is $8 \times \frac{15\sqrt{2}}{7} = \frac{120\sqrt{2}}{7} \approx 24.2$ m.