Phase 1 Consolidation

Q1: Velocity Components (2 marks)

$v_x = v\cos\theta = 12\cos 40° = \mathbf{9.19 \text{ m/s}}$ (1 mark)

$v_y = v\sin\theta = 12\sin 40° = \mathbf{7.71 \text{ m/s}}$ (1 mark)

Q2: Horizontal Launch (2 marks)

(a) Vertical: $s_y = \frac{1}{2}gt^2$ so $t = \sqrt{\frac{2s_y}{g}} = \sqrt{\frac{2 \times 40}{9.8}} = \mathbf{2.86 \text{ s}}$ (1 mark)

(b) Horizontal: $s_x = v_x t = 15 \times 2.86 = \mathbf{42.9 \text{ m}}$ (1 mark)

Q3: Assumptions (2 marks)

Any two of: (1) No air resistance / negligible air resistance; (2) Acceleration due to gravity is constant ($g = 9.8$ m s² downward); (3) Launch and landing heights are the same (level ground); (4) Earth is flat over the trajectory (g is uniform, not radial). (1 mark each)

Q4: Kick on Level Ground (3 marks)

$u_y = 22\sin 35° = 12.62$ m/s; $v_x = 22\cos 35° = 18.02$ m/s

(a) $h_{\max} = \frac{u_y^2}{2g} = \frac{12.62^2}{2 \times 9.8} = \mathbf{8.13 \text{ m}}$ (1 mark)

(b) $t_{\text{flight}} = \frac{2u_y}{g} = \frac{2 \times 12.62}{9.8} = \mathbf{2.58 \text{ s}}$ (1 mark)

(c) $R = v_x \times t = 18.02 \times 2.58 = \mathbf{46.5 \text{ m}}$ (1 mark)
Or use $R = \frac{v^2\sin(2\theta)}{g} = \frac{22^2 \times \sin 70°}{9.8} = 46.5$ m

Q5: Cliff Projectile (3 marks)

$u_y = 18\sin 30° = 9.0$ m/s; $v_x = 18\cos 30° = 15.59$ m/s

Vertical: $s_y = u_y t + \frac{1}{2}at^2$ with $s_y = -25$ m (downward positive convention or $a = -9.8$ with up positive):

$-25 = 9.0t + \frac{1}{2}(-9.8)t^2$ → $-25 = 9.0t - 4.9t^2$

$4.9t^2 - 9.0t - 25 = 0$

Using quadratic formula: $t = \frac{9.0 + \sqrt{81 + 490}}{9.8} = \frac{9.0 + 23.85}{9.8} = \mathbf{3.35 \text{ s}}$ (positive root) (2 marks for method + time)

Horizontal: $s_x = 15.59 \times 3.35 = \mathbf{52.2 \text{ m}}$ (1 mark)

Q6: Complementary Angles (3 marks)

For θ = 20°: $R_1 = \frac{v^2\sin(40°)}{g}$ (1 mark)

For θ = 70°: $R_2 = \frac{v^2\sin(140°)}{g}$ (1 mark)

Since $\sin(140°) = \sin(180° - 40°) = \sin(40°)$, we have $R_1 = R_2$. (1 mark)

Key identity: $\sin(2\theta) = \sin(180° - 2\theta)$, so angles θ and $(90° - \theta)$ give the same range.

Q7: Clearing a Wall (4 marks)

At the wall: $s_x = 15$ m, $s_y = 5$ m, θ = 25°.

$s_x = v\cos\theta \cdot t$ → $t = \frac{15}{v\cos 25°}$ (1 mark)

$s_y = v\sin\theta \cdot t - \frac{1}{2}gt^2$

$5 = v\sin 25° \cdot \frac{15}{v\cos 25°} - 4.9\left(\frac{15}{v\cos 25°}\right)^2$

$5 = 15\tan 25° - \frac{4.9 \times 225}{v^2\cos^2 25°}$

$5 = 6.99 - \frac{1102.5}{v^2 \times 0.821}$

$\frac{1102.5}{0.821v^2} = 6.99 - 5 = 1.99$ (1 mark)

$v^2 = \frac{1102.5}{0.821 \times 1.99} = 675$ → $v = \mathbf{26.0 \text{ m/s}}$ (2 marks)

Q8: Ball to Basket (4 marks)

Given: launch height = 1.5 m, basket at 20 m horizontal, 3.0 m height. So $\Delta s_x = 20$ m, $\Delta s_y = +1.5$ m.

$s_x = v\cos\theta \cdot t = 20$ ... (1)

$s_y = v\sin\theta \cdot t - 4.9t^2 = 1.5$ ... (2) (1 mark)

From (1): $v\cos\theta = \frac{20}{t}$. From (2): $v\sin\theta = \frac{1.5 + 4.9t^2}{t}$

$\tan\theta = \frac{v\sin\theta}{v\cos\theta} = \frac{1.5 + 4.9t^2}{20}$ (1 mark)

Also $v^2 = (v\cos\theta)^2 + (v\sin\theta)^2 = \frac{400}{t^2} + \frac{(1.5 + 4.9t^2)^2}{t^2}$

By trial or solving: try $t = 1.5$ s: $\tan\theta = \frac{1.5 + 11.0}{20} = 0.625$ → θ = 32.0°.

$v = \frac{20}{1.5\cos 32°} = \frac{20}{1.27} = 15.7$ m/s. Check: $s_y = 15.7\sin 32° \times 1.5 - 4.9(1.5)^2 = 8.32 \times 1.5 - 11.0 = 12.5 - 11.0 = 1.5$ m ✓ (2 marks)

θ ≈ 32°, v ≈ 15.7 m/s (accept reasonable ranges from rounding).

Q9: Comparing Projectiles A and B (4 marks)

Given: same speed $v$, A at 30°, B at 60°.

Maximum height: $h \propto \sin^2\theta$. $\sin^2 60° = 0.75$, $\sin^2 30° = 0.25$. So B reaches 3× the height of A. (1 mark)

Flight time: $t \propto \sin\theta$. $\sin 60° = 0.866$, $\sin 30° = 0.5$. So B stays in the air 1.73× longer than A. (1 mark)

Range: $R \propto \sin(2\theta)$. $\sin 60° = \sin 120° = \frac{\sqrt{3}}{2}$. Same range. (1 mark)

Explanation: 30° and 60° are complementary angles. The smaller angle gives less vertical velocity (shorter time, lower height) but more horizontal velocity. The larger angle gives more vertical velocity (longer time, greater height) but less horizontal velocity. These two effects exactly balance for range because $\sin(2\theta)$ is symmetric about 45°. (1 mark)

Q10: Firework Problem (5 marks)

Let time to max height = $t_1$, time from explosion to landing = $t_2$, total time = 4.5 s.

At max height: spark travels horizontally at 5 m/s. Horizontal distance from explosion point = $5t_2$.

Total horizontal distance from launch = $v\cos\theta \cdot t_1 + 5 \cdot t_2 = 45$ m ... (1) (1 mark)

Time to max height: $t_1 = \frac{v\sin\theta}{g}$. Explosion height: $h = \frac{(v\sin\theta)^2}{2g}$.

Fall time from height $h$: $h = \frac{1}{2}gt_2^2$ so $t_2 = \sqrt{\frac{2h}{g}} = \frac{v\sin\theta}{g} = t_1$. (1 mark)

So $t_1 = t_2 = \frac{4.5}{2} = 2.25$ s. (1 mark)

From $t_1 = \frac{v\sin\theta}{g}$: $v\sin\theta = 9.8 \times 2.25 = 22.05$ m/s.

From (1): $v\cos\theta \times 2.25 + 5 \times 2.25 = 45$ → $2.25(v\cos\theta + 5) = 45$

$v\cos\theta + 5 = 20$ → $v\cos\theta = 15$ m/s. (1 mark)

$v = \sqrt{(v\sin\theta)^2 + (v\cos\theta)^2} = \sqrt{22.05^2 + 15^2} = \sqrt{486.2 + 225} = \sqrt{711.2} = \mathbf{26.7 \text{ m/s}}$

$\tan\theta = \frac{22.05}{15} = 1.47$ → $\theta = \mathbf{55.8°}$ (1 mark)

Q11: Long Jumper (4 marks)

(a) $v_x = 9.2\cos 22° = \mathbf{8.53 \text{ m/s}}$; $u_y = 9.2\sin 22° = \mathbf{3.45 \text{ m/s}}$ (1 mark)

(b) $t = \frac{2u_y}{g} = \frac{2 \times 3.45}{9.8} = \mathbf{0.704 \text{ s}}$ (1 mark)

(c) $R = v_x \times t = 8.53 \times 0.704 = \mathbf{6.01 \text{ m}}$ (1 mark)

(d) $h_{\max} = \frac{u_y^2}{2g} = \frac{3.45^2}{19.6} = \mathbf{0.607 \text{ m}}$ (1 mark)

Q12: Building Throw (4 marks)

$v_x = 15\cos 50° = 9.64$ m/s; $u_y = 15\sin 50° = 11.49$ m/s

(a) Time to max height: $t_{up} = \frac{u_y}{g} = \frac{11.49}{9.8} = \mathbf{1.17 \text{ s}}$ (1 mark)

(b) For total flight: $s_y = -30$ m (down). $-30 = 11.49t - 4.9t^2$ → $4.9t^2 - 11.49t - 30 = 0$

$t = \frac{11.49 + \sqrt{132.0 + 588}}{9.8} = \frac{11.49 + 26.93}{9.8} = \mathbf{3.92 \text{ s}}$ (1 mark)

(c) $s_x = 9.64 \times 3.92 = \mathbf{37.8 \text{ m}}$ (1 mark)

(d) At impact: $v_x = 9.64$ m/s (constant). $v_y = u_y - gt = 11.49 - 9.8(3.92) = 11.49 - 38.4 = -26.9$ m/s.

$v = \sqrt{9.64^2 + 26.9^2} = \mathbf{28.6 \text{ m/s}}$. Direction: $\tan\theta = \frac{26.9}{9.64}$ below horizontal, θ = 70.3°. (1 mark)

Q13: Two Possible Angles (4 marks)

(a) $R = \frac{v^2\sin(2\theta)}{g}$ → $72 = \frac{30^2 \sin(2\theta)}{9.8}$

$\sin(2\theta) = \frac{72 \times 9.8}{900} = 0.784$ (1 mark)

$2\theta = \sin^{-1}(0.784) = 51.6°$ or $180° - 51.6° = 128.4°$

$\theta = \mathbf{25.8°}$ or $\mathbf{64.2°}$ (1 mark)

(b) For θ = 25.8°: $t = \frac{2 \times 30 \times \sin 25.8°}{9.8} = \frac{60 \times 0.435}{9.8} = \mathbf{2.66 \text{ s}}$

For θ = 64.2°: $t = \frac{2 \times 30 \times \sin 64.2°}{9.8} = \frac{60 \times 0.901}{9.8} = \mathbf{5.52 \text{ s}}$ (1 mark)

(c) There are two angles because $\sin(2\theta)$ has the same value for $2\theta$ and $(180° - 2\theta)$. The lower angle gives a smaller vertical component (shorter time, lower max height) but larger horizontal component. The higher angle gives a larger vertical component (longer time, greater max height) but smaller horizontal component. These two combinations produce the same range because the extra flight time of the high angle exactly compensates for the reduced horizontal speed. (1 mark)

MCQ Answers

1. B ($20\sin 60° = 17.3$ m/s)

2. C (At max height, $v_y = 0$ only)

3. B (Launch = landing height required)

4. A (Doubling $v$ doubles $R$ since $R \propto v$)

5. B (30° and 60° are complementary → same range)

SA1: Assessing the Range Claim (3 marks)

The claim is incorrect. From $R = \frac{v^2\sin(2\theta)}{g}$, range depends on $\sin(2\theta)$, which is maximum at $\theta = 45°$. (1 mark) Increasing the angle from, say, 30° to 45° does increase range. But increasing beyond 45° decreases range because $\sin(2\theta)$ decreases. (1 mark) For example, $\sin(60°) = 0.866$ at 30° and $\sin(90°) = 1.0$ at 45°, but $\sin(120°) = 0.866$ at 60° — same range as 30°. So range increases up to 45° then decreases. (1 mark)

SA2: Projectile at 16 m/s, 40° (3 marks)

$v_x = 16\cos 40° = \mathbf{12.3 \text{ m/s}}$; $u_y = 16\sin 40° = \mathbf{10.3 \text{ m/s}}$ (1 mark)

$t_{up} = \frac{u_y}{g} = \frac{10.3}{9.8} = \mathbf{1.05 \text{ s}}$ (1 mark)

$h_{\max} = \frac{u_y^2}{2g} = \frac{10.3^2}{19.6} = \mathbf{5.41 \text{ m}}$ (1 mark)

SA3: Evaluate Time of Flight Statement (4 marks)

The statement is correct. (1 mark) The flight time is found from vertical motion only. For level ground: $t = \frac{2v\sin\theta}{g} = \frac{2u_y}{g}$, which depends only on $u_y$, the vertical component. (1 mark) The horizontal component $v_x = v\cos\theta$ does not appear in the flight time equation. Numerical example: launch at 20 m/s at 30° vs 60°. Same $v$, different flight times: $t_{30} = \frac{2 \times 20 \times 0.5}{9.8} = 2.04$ s; $t_{60} = \frac{2 \times 20 \times 0.866}{9.8} = 3.54$ s. The horizontal component changed (17.3 m/s vs 10 m/s) but flight time changed because the vertical component changed, not the horizontal. (1 mark) Even if $v_x$ were zero (straight up), time depends only on $u_y$. If $v_x$ doubled while $u_y$ stayed the same, flight time would not change. (1 mark)