1.

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 :

Answer»

`sqrt(H/g)`
`sqrt((2H)/g)`
`sqrt(H/(2G))`
`sqrt((2H)/(gcosbeta))`

SOLUTION :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)`


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