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Self-check: Projectile motion

How this works

Type your answer and press Check. Nothing is saved or sent anywhere; this is just for you. Stuck? Open the hint first, then the worked solution. Use \(g = 9.8\) m s⁻², ignore air resistance, and give answers to 3 significant figures. Checklist codes: G1.1 G1.3 G1.4 G1.5

Notes for this topic: Projectile motion

1. Components of a launch velocity

G1.1

A ball is launched at 18.0 m s⁻¹ at 35° above the horizontal.

(a) Horizontal component of the launch velocity, \(u_x\):

(b) Vertical component of the launch velocity, \(u_y\):

Hint

Sketch the velocity as the hypotenuse of a right triangle. The side next to the 35° angle is horizontal.

Worked solution

Given: \(u = 18.0\) m s⁻¹, \(\theta = 35^\circ\)
Formula: \(u_x = u\cos\theta\), \(u_y = u\sin\theta\)
Substitute: \(u_x = 18.0\cos 35^\circ = 14.7\) m s⁻¹; \(u_y = 18.0\sin 35^\circ = 10.3\) m s⁻¹
Check: \(\sqrt{14.7^2 + 10.3^2} = 18.0\) ✓

2. Dropped vs launched

G1.3

From the same height, ball A is dropped and ball B is launched horizontally at 5 m s⁻¹ at the same instant. Which lands first?

3. Horizontal launch

G1.4

A marble rolls off a bench 0.95 m high at 2.4 m s⁻¹. How far from the foot of the bench does it land?

Hint

Find the time to fall 0.95 m using the vertical motion only (\(u_y = 0\)). Then use \(s_x = u_x t\).

Worked solution

Vertical: \(s_y = \tfrac{1}{2} g t^2 \Rightarrow t = \sqrt{\dfrac{2(0.95)}{9.8}} = 0.440\) s
Horizontal: \(s_x = u_x t = 2.4 \times 0.440 = 1.06\) m
Check: a slow marble off a bench landing about 1 m away is reasonable.

4–6. Kicked from the ground

G1.4

A ball is kicked at 22 m s⁻¹ at 40° above the horizontal from level ground, and lands back on level ground.

4. Time of flight:

5. Horizontal range:

6. Maximum height:

Hint

Resolve first: \(u_x = 22\cos 40^\circ\), \(u_y = 22\sin 40^\circ\). At maximum height \(v_y = 0\).

Worked solution

Components: \(u_x = 22\cos 40^\circ = 16.85\) m s⁻¹, \(u_y = 22\sin 40^\circ = 14.14\) m s⁻¹
4. Time to top: \(0 = u_y - g t \Rightarrow t_{top} = 14.14/9.8 = 1.443\) s, so the flight time is \(2 \times 1.443 = 2.89\) s
5. Range: \(s_x = u_x t = 16.85 \times 2.886 = 48.6\) m
6. Max height: \(0 = u_y^2 - 2gh \Rightarrow h = \dfrac{14.14^2}{2(9.8)} = 10.2\) m

7. Using range data

G1.5

A launcher fires a ball at the same speed each time, from level ground. At 25° the range is 6.4 m. Predict the range at 65°.

Hint

On level ground, range depends on \(\sin 2\theta\). Compare \(\sin 50^\circ\) and \(\sin 130^\circ\).

Worked solution

Range \(\propto \sin 2\theta\). \(\sin(2 \times 25^\circ) = \sin 50^\circ\) and \(\sin(2 \times 65^\circ) = \sin 130^\circ = \sin 50^\circ\), so the ranges are equal: 6.4 m. Complementary angles (adding to 90°) give the same range; 45° gives the maximum.

8. Explain it

G1.3

Explain why the horizontal velocity of a projectile stays constant during its flight. Write your answer, then compare it with the marking points.

Marking points (self-mark)
  • Ignoring air resistance, the only force on the projectile is gravity (its weight). ✔
  • Gravity acts vertically, so there is no horizontal force. ✔
  • With no horizontal net force there is no horizontal acceleration (Newton's first or second law), so \(v_x\) stays constant. ✔

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