Answer to Question #81915 in Mechanics | Relativity for yashwanth

Question

A projectile is thrown at angle β 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(t)=v_0 sin⁡〖β t〗

y(t)=v_0 cos⁡〖β t〗-(gt^2)/2

The total time of motion

t=(2v_0 cos⁡β)/g

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

t_H=t/2=(v_0 cos⁡β)/g

The maximum height

H=y(t_H )=(v_0^2 cos^2⁡β)/2g=(gt_H^2)/2

Therefore

t_H=√(2H/g)