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The horizontal range of a projectile is ...

The horizontal range of a projectile is `2 sqrt(3)` times its maximum height. Find the angle of projection.

Text Solution

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If u and `alpha` be the initial velocity of projection and anggle of projection respectively, then the maximum height attained `H_(m)=(u^(2)sin^(2)alpha)/(2g)` and horizontal range `R=(2u^(2)sin alpha cos alpha)/g`
According to the problem we can write
`(2u^(2)sin alpha cos alpha)/g=2sqrt(3)((u^(2)sin^(2)alpha)/(2g))impliestan alpha=(2/(sqrt(3)))impliesalpha=tan^(-1)(2/(sqrt(3)))`
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