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

At the maximum height $H$, we can use the equation:

$H = v_0 T - \frac{1}{2} g T^2$

Substituting $v_0 = u$ and $T = \frac{u}{g}$:

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

This simplifies to:

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

Let the total time to hit the ground be $T_{ ext{total}}$. The height of point $A$ relative to the ground is $3H$.

The distance fallen is: $AH = 3H$

Using the equation of motion $s = ut + \frac{1}{2} a t^2$ with initial velocity $u$ and acceleration $-g$, we set up the equation:

$-3H = uT_{ ext{total}} - \frac{1}{2} g T_{ ext{total}}^2$

Substituting $H = \frac{u^2}{2g}$ gives:

$-3\left(\frac{u^2}{2g}\right) = uT_{ ext{total}} - \frac{1}{2} g T_{ ext{total}}^2$

Solving this leads to: $T_{ ext{total}} = T + 2T$

Thus, $T_{ ext{total}} = 3T$