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

For particle B, which is released from a height of $kh$, we again use the equation of motion:

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

Here, $u = 0$, and it is released after a time of $t$ seconds, so the equation becomes:

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

Substituting the expression for $h$ into the equation for particle B:

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

Simplifying gives:

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

Taking the square root of both sides:

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

Thus,

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

This proves the required relationship:

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

The result is valid within the conditions outlined ($0 < k < 1$).