At time $t = 0$, a particle is projected vertically upwards with speed $u$ from a point A - Edexcel - A-Level Maths Mechanics - Question 4 - 2014 - Paper 2

The maximum height $H$ can be calculated using the kinematic equation:

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

Substituting $T$ into the equation, we get:

$H = uT + \frac{1}{2}(-g)T^2$

Plugging in $T = \frac{u}{g}$:

$H = u\left(\frac{u}{g}\right) + \frac{1}{2}(-g)\left(\frac{u}{g}\right)^2$

This simplifies to:

$H = \frac{u^2}{g} - \frac{1}{2}\frac{gu^2}{g^2} = \frac{u^2}{g} - \frac{u^2}{2g} = \frac{u^2}{2g}.$

From part (b), we found $H = \frac{u^2}{2g}$. Given that point A is at a height $3H$:

$3H = 3\frac{u^2}{2g} = \frac{3u^2}{2g}.$

The time it takes to fall from height $3H$ to the ground can be calculated using:

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

But since it is falling from rest:

$3H = \frac{1}{2}gt^2$

To find $t$ (the time taken to fall), we substitute for $H$:

$3\frac{u^2}{2g} = \frac{1}{2}gt^2 \implies t^2 = \frac{3u^2}{g} \implies t = \sqrt{\frac{3u^2}{g}} = \frac{u\sqrt{3}}{\sqrt{g}}.$

So, the total time from projection to hitting the ground is:

$T + t = T + \frac{u\sqrt{3}}{\sqrt{g}} = T + 2T = 3T.$