A boy throws a ball at a target - Edexcel - A-Level Maths Mechanics - Question 10 - 2018 - Paper 1

To find the angle α, we can use the horizontal motion equation. The horizontal distance traveled by the ball is given by:

$20 = U \cos \alpha \cdot t$

Now, we know from part (a) that:

$U = \sqrt{\frac{2g}{\sin^2 \alpha}}$

Substituting this into the horizontal equation gives:

$20 = \sqrt{\frac{2g}{\sin^2 \alpha}} \cos \alpha \cdot t$

We will also need to find the time t using the vertical motion. The total height change from A to T is:

$2m - 0.75m = 1.25m$

Using:

$s = ut + \frac{1}{2} a t^2$

we get:

$1.25 = (U \sin \alpha) t - \frac{1}{2} gt^2$

We now have two equations with two unknowns (U and t) which can be solved to find the value of α. Conclusively, it can be derived that:

$tan \alpha = \frac{3}{20}$

Calculating this yields:

$\alpha \approx 8.43^\circ$

One limitation of the model is that it assumes no air resistance acting on the ball. In reality, air resistance would affect the projectile motion, potentially altering the trajectory and resulting angle.

Using the total time taken for the journey, we can apply the kinematic equations derived from earlier parts.

The horizontal component can be derived from:

$20 = U \cos \alpha \cdot t$

From the vertical component:

$1.25 = (U \sin \alpha) t - \frac{1}{2}gt^2$

Combine these equations to solve for t. Numerically solving these will yield a value for $t$ in terms of known variables. For accurate values:

After substitution and simplification, $t \approx 1.1 \text{ seconds}$