Checkpoint 1: Projectile Motion

Multiple Choice Answers

1. A — $v_x = v\cos\theta = 25 \times \cos 40° = 19.2$ m/s. Option B uses sine which gives the vertical component. Option C incorrectly uses tangent. Option D ignores the angle entirely.

2. B — Horizontal velocity remains constant because there is no horizontal acceleration (ignoring air resistance). Only gravity acts, and it acts vertically.

3. C — $t = \frac{2v\sin\theta}{g}$. The factor of 2 appears because the projectile must go up and come back down. Option A gives the time to max height only. Option D is the range equation without the $v$ term.

4. C — $h_{\max} = \frac{(v\sin\theta)^2}{2g} = \frac{(30 \times \sin 60°)^2}{2 \times 9.8} = \frac{(25.98)^2}{19.6} = \frac{675}{19.6} = 34.4$ m.

5. C — Complementary angles give equal range because $\sin(2\theta_1) = \sin(2\theta_2)$ when $\theta_1 + \theta_2 = 90°$. Here $\sin(60°) = \sin(120°) = 0.866$.

6. C — At maximum height, vertical velocity $v_y = 0$ (momentarily), but acceleration is always $g = 9.8$ m/s$^2$ downward. Horizontal velocity remains non-zero.

7. C — Range $R = \frac{v^2\sin(2\theta)}{g}$ is maximised when $\sin(2\theta) = 1$, giving $\theta = 45°$. This assumes launch and landing at the same height.

8. C — $t = \frac{2v\sin\theta}{g} = \frac{2 \times 20 \times 0.6}{10} = \frac{24}{10} = 2.4$ s.

9. C — Thrown horizontally means $u_y = 0$. Using $s_y = \frac{1}{2}gt^2$: $45 = 5t^2$, so $t = \sqrt{9} = 3.0$ s.

10. B — The vertical component is always found using sine: $v_y = v\sin\theta$. The horizontal component uses cosine.

11. B — Since $R = \frac{v^2\sin(2\theta)}{g}$, range is inversely proportional to $g$. Halving $g$ doubles the range.

12. B — 20° and 70° are complementary ($20° + 70° = 90°$), so $\sin(40°) = \sin(140°)$ and the ranges are equal. The 70° projectile reaches greater height and has longer flight time.

13. B — $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 80}{10}} = \sqrt{16} = 4$ s. Then $s_x = v_x \times t = 5 \times 4 = 20$ m.

14. D — Mass does not appear in any projectile motion equation. With no air resistance, all objects fall at the same rate regardless of mass (Galileo's principle).

15. A — Taking upward as positive, the displacement is $-20$ m (ground is 20 m below launch). Initial vertical velocity is $v\sin\theta = 15\sin30°$. Using $s_y = u_y t + \frac{1}{2}at^2$ with $a = -g = -10$ m/s$^2$ gives $-20 = (15\sin30°)t - 5t^2$.

Short Answer 1 — Kicked Ball (3 marks)

(a) Components of initial velocity (1 mark)

$v_x = v\cos\theta = 18 \times \cos 35° = 18 \times 0.819 = 14.7$ m/s

$v_y = v\sin\theta = 18 \times \sin 35° = 18 \times 0.574 = 10.3$ m/s

(b) Maximum height (1 mark)

$h_{\max} = \frac{(v\sin\theta)^2}{2g} = \frac{(10.3)^2}{2 \times 9.8} = \frac{106.1}{19.6} = 5.41$ m

Alternatively, using $v_y^2 = u_y^2 + 2as_y$ with $v_y = 0$:

$0 = (10.3)^2 + 2(-9.8)h_{\max}$, so $h_{\max} = \frac{106.1}{19.6} = 5.41$ m

(c) Range (1 mark)

$t_{\text{flight}} = \frac{2v\sin\theta}{g} = \frac{2 \times 10.3}{9.8} = 2.10$ s

$R = v_x \times t = 14.7 \times 2.10 = 30.9$ m

Alternatively using range equation: $R = \frac{v^2\sin(2\theta)}{g} = \frac{18^2 \times \sin 70°}{9.8} = \frac{324 \times 0.940}{9.8} = 31.1$ m (small difference due to rounding)

Short Answer 2 — Stone from Cliff (3 marks)

(a) Time to reach the sea (1 mark)

Thrown horizontally: $u_y = 0$, $s_y = 50$ m, $a = g = 9.8$ m/s$^2$

$s_y = \frac{1}{2}gt^2$

$50 = 4.9t^2$

$t = \sqrt{\frac{50}{4.9}} = \sqrt{10.2} = 3.19$ s $\approx$ 3.2 s

(b) Horizontal distance (1 mark)

$s_x = v_x \times t = 10 \times 3.19 = 31.9$ m $\approx$ 32 m

(c) Speed at impact (1 mark)

$v_x = 10$ m/s (constant)

$v_y = u_y + gt = 0 + 9.8 \times 3.19 = 31.3$ m/s

$v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 31.3^2} = \sqrt{100 + 979.7} = \sqrt{1079.7} = 32.9$ m/s $\approx$ 33 m/s

Short Answer 3 — Complementary Angles (4 marks)

Showing equal range (2 marks)

For 25°: $R = \frac{v^2\sin(2\theta)}{g} = \frac{25^2 \times \sin(50°)}{9.8} = \frac{625 \times 0.766}{9.8}$

For 65°: $R = \frac{25^2 \times \sin(130°)}{9.8} = \frac{625 \times 0.766}{9.8}$

Since $\sin(50°) = \sin(130°) = 0.766$ (because $50° + 130° = 180°$, and $\sin(180° - \alpha) = \sin\alpha$), the ranges are equal.

Alternatively, note that 25° and 65° are complementary: $25° + 65° = 90°$. For complementary angles, $\sin(2\theta_1) = \sin(2\theta_2)$ because $2\theta_1 + 2\theta_2 = 180°$.

Calculating the range (2 marks)

$R = \frac{625 \times 0.7660}{9.8} = \frac{478.8}{9.8} = 48.9$ m

The range for both launch angles is approximately 48.9 m (accept 49 m).

Short Answer 4 — Clearing the Wall (4 marks)

Setting up the problem (1 mark)

We need the ball to reach the wall ($s_x = 12$ m) at a height of at least 3.5 m. Launch height is 1.8 m, so vertical displacement needed: $\Delta h = 3.5 - 1.8 = 1.7$ m.

Let the launch speed be $v$. Then $v_x = v\cos 40°$ and $v_y = v\sin 40°$.

Finding time to reach the wall (1 mark)

$t = \frac{s_x}{v_x} = \frac{12}{v\cos 40°}$

Vertical motion equation (1 mark)

$s_y = v_y t + \frac{1}{2} a t^2$

$1.7 = (v\sin 40°) \times \frac{12}{v\cos 40°} + \frac{1}{2}(-9.8) \times \left(\frac{12}{v\cos 40°}\right)^2$

$1.7 = 12\tan 40° - \frac{4.9 \times 144}{v^2 \cos^2 40°}$

$1.7 = 12 \times 0.839 - \frac{705.6}{v^2 \times 0.587}$

$1.7 = 10.07 - \frac{1202}{v^2}$

Solving for $v$ (1 mark)

$\frac{1202}{v^2} = 10.07 - 1.7 = 8.37$

$v^2 = \frac{1202}{8.37} = 143.6$

$v = \sqrt{143.6} = 12.0$ m/s

The minimum launch speed required is approximately 12 m/s.

Short Answer 5 — Mass and Projectile Motion (4 marks)

Evaluation (1 mark)

The statement is incorrect. In the absence of air resistance, mass has no effect on projectile motion.

Physics principle (1 mark)

The equations of projectile motion contain no mass term. The acceleration of any projectile in a gravitational field is $g = 9.8$ m/s$^2$, independent of mass. This is a consequence of the equivalence of inertial and gravitational mass.

Mathematical justification (1 mark)

Range: $R = \frac{v^2\sin(2\theta)}{g}$ — no mass term.

Time of flight: $t = \frac{2v\sin\theta}{g}$ — no mass term.

Maximum height: $h_{\max} = \frac{(v\sin\theta)^2}{2g}$ — no mass term.

Conclusion (1 mark)

Since all projectile motion depends only on initial velocity, launch angle, and $g$ (not mass), two projectiles of different masses launched at the same speed and angle will follow identical trajectories and travel the same distance. This was famously demonstrated by Galileo and confirmed by the Apollo astronauts who dropped a hammer and a feather on the Moon (where there is no air), and they fell at the same rate.