Derive the expressions for the time taken to reach maximum height,the total time of flight,and the maximum height attained by a projectile.

(N/A) $1$. Time taken to reach maximum height $(t_m)$:
At maximum height,the vertical component of velocity $(v_y)$ is zero.
Using the equation $v_y = v_0 \sin \theta_0 - gt$,where $v_y = 0$ at $t = t_m$:
$0 = v_0 \sin \theta_0 - gt_m$
$t_m = \frac{v_0 \sin \theta_0}{g}$
$2$. Total time of flight $(T_f)$:
The total time of flight is the time taken to return to the ground $(y = 0)$.
Using the displacement equation $y = (v_0 \sin \theta_0)t - \frac{1}{2}gt^2$:
$0 = (v_0 \sin \theta_0)T_f - \frac{1}{2}gT_f^2$
$T_f = \frac{2v_0 \sin \theta_0}{g}$
$3$. Maximum height $(H)$:
At maximum height,$t = t_m = \frac{v_0 \sin \theta_0}{g}$.
Substituting this into the vertical displacement equation:
$H = (v_0 \sin \theta_0)t_m - \frac{1}{2}gt_m^2$
$H = (v_0 \sin \theta_0) \left( \frac{v_0 \sin \theta_0}{g} \right) - \frac{1}{2}g \left( \frac{v_0 \sin \theta_0}{g} \right)^2$
$H = \frac{v_0^2 \sin^2 \theta_0}{g} - \frac{v_0^2 \sin^2 \theta_0}{2g}$
$H = \frac{v_0^2 \sin^2 \theta_0}{2g}$