1.

A projectile is projected with velocity V such that its range is twice the greatest height attained, the correct value of its range is :

Answer»

`V^(2)/g`
`3/5 V^(2)/g`
`4/5V^(2)/g`
`1/5 V^(2)/g`

Solution :R=2Hbr>`(2V^(2)sinthetacostheta)/g=2.(v^(2)sin^(2)theta)/(2g)`
`implies tantheta=2/1`
`:. Sintheta=2/sqrt(5)` and `costheta=1/sqrt(5)`
`:. R=(2v^(2)xx2/sqrt(5) xx 1/sqrt(5))/g=(4v^(2))/(5g)`


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