A projectile is thrown at angel with vertical. It reaches a maximum height H. The time taken to reach the highest point of its path is :

Options

A. `sqrt(H/g)`
B. `sqrt((2H)/g)`
C. `sqrt(H/(2g))`
D. `sqrt((2H)/(gcosbeta))`

Answer

B

Explanation

Here `(v^(2)cos^(2)beta)/(2g)=H:.vcosbeta=sqrt(2gH)`
But the time of flight is
`t=(vcosbeta)/g=sqrt(2gh)/g` or `t=sqrt((2H)/g)`