1.

In case of a projectile fired at an angle equally inclined to a horizontal and vertical with velocity v, the greatest height is given by :

Answer»

`v^(4)/(4G)`
`v^(2)/(4g)`
`(2v^(2))/4`
`v^(2)/g^(2)`

SOLUTION :`H=(v^(3)sin^(2)theta)/(2g)=v^(2)/(2g)sin^(2)(45^@)=v^(2)/(4g)`


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