Question

A particle is projected vertically upwards from a point A on the ground$.$ it makes time $t_1$ to reach a point B$,$ but it still continues to move up $.$ If it takes further time $t_2$ to reach the ground from point B$.$ Then height of point B from the ground is$:$

Answer 1

Let the height of point B be 'h' and initial velocity of particle be 'u'.

Total time of light $=t_1+t_2$

So, time to reach at maximum height $=\frac{t_1+t_2}{2}$

Applying I equation of motion

$0=u+(-g)\left[\frac{t_1+t_2}{2}\right]$

$\Rightarrow u=\frac{g(t_1+t_2)}{2}$

Applying II equation of motion

$h=u{t}_1-\frac{1}{2}g{t}_1^2$

$=\frac{g(t_1+t_2)t_1}{2}-\frac{g{t}_1^2}{2}$

$\Rightarrow h=\frac{g{t}_1{t}_2}{2}$.