.Time taken to reach the maximum height 'H' be $t_{\max }$

At maximum height the y component of the velocity (vg) of a projectile is zero, we get,

$y_{y}=v_{\mathrm{o}} \sin \theta_{\mathrm{o}}-g t$

$y_{\mathrm{y}}=v_{\mathrm{o}} \sin \theta_{\mathrm{o}}-g t$

$\therefore 0=v_{\mathrm{o}} \sin \theta_{\mathrm{o}}-g t_{m}$

$\therefore g t_{m}=v_{\mathrm{o}} \sin \theta_{\mathrm{o}}$

$\therefore t_{m}=\frac{v_{o} \sin \theta_{o}}{g}$

Total time of flight is given by

Let it be $t_{p}$

substituting $y=\left(v_{0} \sin \theta_{\mathrm{o}}\right) t-1 / 2 g t^{2}$ when $y=0$ then $t=t_{\mathrm{F}}$

$\therefore 0=\left(v_{0} \sin \theta_{o}\right) t_{\mathrm{F}}-1 / 2 g t^{2} \mathrm{~F}$

$\therefore 0=v_{0} \sin \theta_{o}-1 / 2 g t_{\mathrm{F}}$

$\therefore t_{\mathrm{F}}=\frac{2 v_{o} \sin \theta_{o}}{g}$

Maximum height in projectile motion :

Maximum Height attained is shown in figure.

The distance travelled by the body in vertical direction is given by $y=\left(v_{0} \sin \theta_{0}\right) t-1 / 2 g t^{2}$

Putting $t=t_{m}$ we have then $y=\mathrm{H}$.

$\therefore \mathrm{H}=\left(v_{0} \sin \theta_{0}\right) t_{m}-1 / 2 g t_{m}^{2}$

Substituting the value of $\mathrm{t}_{m}$ from (1) we get,

$\therefore \mathrm{H}=\left[v_{\mathrm{o}} \sin \theta_{\mathrm{o}}\right] t_{m}-\frac{1}{2} g t_{m}^{2}$