Time taken by the body to reach the point $\mathrm{A}$ is $t_1$ (During upward journey).



The body crosses this point again (during the downward journey) after $t_2, ie,$the body takes the time $\left(t_2-t_1\right)$ to come again at the point $\mathrm{A}.$

So, the time taken by the body to reach at point $B$ (a maximum height).

$t=t_1+\left(\frac{t_2-t_1}{2}\right)$

$\left[\because \mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{g}=\mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{d}\mathrm{e}\mathrm{s}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{g}\right]$

$t=\frac{t_1+t_2}{2}$

So, maximum height $H=\frac{1}{2}\mathrm{ }\mathrm{g}{t}^2$

$=\frac{1}{2}\mathrm{g}{\left(\frac{t_1+t_2}{2}\right)}^2$

$=2\mathrm{g}{\left(\frac{t_1+t_2}{4}\right)}^2$