Question

A projectile is thrown at angle $\beta$ with vertical. It reaches a maximum height $H$. The time taken to reach the highest point of its path is?

Answer 1

The projectile equations of motion are as follows

$x\left(t\right)=v_0\sin \left[\beta t\right]$

$y\left(t\right)=v_0\cos \left[\beta t\right]-\frac{\left(g{t}^2\right)}{2}$

The total time of motion

$t=\frac{\left(2v_0\cos \beta \right)}{g}$

The time taken to reach the highest point of its path is:

$t_H=\frac{t}{2}=\frac{\left(v_0\cos \beta \right)}{g}$

$H=y\left(t_H\right)=\frac{\left(v_0^2{\cos }^2\beta \right)}{2g}=\frac{\left(g{t}_H^2\right)}{2}$

$t_H=\sqrt{\frac{2H}{g}}$