A body projected vertically upwards crosses a point twice in its journey at a height $h$ just after $t_1$…
A body projected vertically upwards crosses a point twice in its journey at a height $h$ just after $t_1$ and $t_2$ seconds. Maximum height reached by the body is
$\frac{g}{4}\left(t_1+t_2\right)^2$
$g\left(\frac{t_1+t_2}{4}\right)^2$
$2 g\left(\frac{t_1+t_2}{4}\right)^2$
$\frac{g}{4}\left(t_1 t_2\right)$
Solution
Time taken by the body to reach the point $A$ is $t_1$ (During upward journey).
The body crosess this point again (during downward journey) after $t_2$, i.e., the body takes the time $\left(t_2-t_1\right)$ to come again at point $A$.
So, the time taken by the body to reach at point $B$ (at maximum height).
$t=t_1+\left(\frac{t_2-t_1}{2}\right)$
[ $\because$ Time of ascending $=$ Time of descending]
$t=\frac{t_1+t_2}{2}$
So, maximum height, $H=\frac{1}{2} g t^2=\frac{1}{2} g\left(\frac{t_1+t_2}{2}\right)^2$
$=2 g\left(\frac{t_1+t_2}{4}\right)^2$