3. (a) A particle is projected with a speed of 98 m s⁻¹ at an angle α to the horizontal - Leaving Cert Applied Maths - Question 3 - 2012

We use the time of flight formula for projectile motion:

$t_f = \frac{2v \sin(α)}{g}$

Calculating times for both angles:

• For $α_1 \approx 39.4°$: $t_{f1} = \frac{2 \times 98 \times \sin(39.4°)}{9.8} \approx 12.00 \text{ s}$

• For $α_2 \approx 140.6°$: $t_{f2} = \frac{2 \times 98 \times \sin(140.6°)}{9.8} \approx 16.00 \text{ s}$

Thus, the difference in time of flight is: $t_f = t_{f2} - t_{f1} = 16.00 - 12.00 = 4.00 ext{ s}$

For the inclined plane, we analyze the motion as follows: Using the time of flight for the particle projected down the incline:

$r_f = 0$ $20t - \frac{1}{2}g \sin(α)t^2 = 0$

From here we can deduce:

• Solving for time $t$ gives us: $t = \frac{40}{g \cos(α)}$

Using the provided range on the inclined plane: $r_f = \frac{1600 \sqrt{3}}{g}$

• Setting up the equation: $0 = \frac{1600 \sqrt{3}}{g}$

With this we can deduce: $\frac{1}{2}g \sin(α) \cdot \frac{1600 \sqrt{3}}{g^2} = \frac{1600 \sqrt{3}}{8}$

• Solving yields: $\sin(α) = \frac{2 \sqrt{3}}{3}.$
• Thus, the angle is: $α \approx 60°$.

To find the rebound velocity, we use the equations derived from the conservation of momentum and the coefficient of restitution:

Let $v_r$ be the rebound velocity and $v_i$ be the incoming velocity:

• Using the relation for coefficient of restitution: $e = \frac{|v_r|}{|v_i|}$
• Given that $e = \frac{1}{2}$ and substituting values: $|v_r| = \frac{1}{2} |v_i|$

Calculating the incoming velocity at Q:

• From earlier calculations: $v_i = \sqrt{(40 \sqrt{3})^2 + (10)^2}$
• Solving gives: $|v_i| = 70 \text{ m/s},$
• Then, the rebound velocity: $|v_r| = \frac{1}{2} \times 70 = 35 \text{ m/s}$.