Two particles A and B are released from rest from different starting points above a horizontal surface - AQA - A-Level Maths Mechanics - Question 16 - 2020 - Paper 2

For particle B, released at time (t), we again use the equation of motion:

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

Here, the initial velocity (u = 0), acceleration (a = g), and it travels for (5 - t) seconds. Hence:

$kh = \frac{1}{2} g (5 - t)^2$

Substituting (h = \frac{25g}{2}) gives:

$k \left( \frac{25g}{2} \right) = \frac{1}{2} g (5 - t)^2$

Rearranging gives:

$25k = (5 - t)^2$

Taking square roots:

$5 \sqrt{k} = 5 - t$

Therefore,

t = 5 - 5 \sqrt{k}\Rightarrow t = 5(1 - \sqrt{k})$$

Thus, we have proven:

t = 5(1 - \sqrt{k})\text{, as required.}