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 from a height of $kh$ at a time $t$ seconds after particle A:

Using the same equation of motion:

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

Here, particle B is released after $t$ seconds, so we set the total time as $5 - t$ seconds:

$s_B = 0 \cdot (5 - t) + \frac{1}{2}g(5 - t)^2 = \frac{g(5 - t)^2}{2}$

Since both particles land on the same surface:

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

From the heights we established:

$kh = \frac{25g}{2} (5 - t)^2$

Substituting $h$:

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

Multiplying through by 2 and cancelling $g$ gives:

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

Taking the square root:

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

Thus,

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

Rearranging leads us to:

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