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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.

A

` tan^-1 (sqrt3/(sqrt(2)))`

B

` sin^-1 ((2)/(sqrt(3)))`

C

` cot^-1 ((2)/(sqrt(3)))`

D

` tan^-1 ((2)/(sqrt(3)))`

Text Solution

Verified by Experts

If `u and prop` are the initial velocity of projection and angle of projection, respectively, then
Maximum height attained `=(u^2 sin^2 prop)/(2 g)`
Horizontal range `(2 u^2 sin prop cos prop)/(g)`
According to the problem,
`(2 u^2 sin prop cos prop)/(g) = 2 sqrt(3) ((u^2 sin^2 prop)/(2 g)) rArr tan prop = ((2)/(sqrt(3)))`
`rArr prop = tan^-1 ((2)/(sqrt(3)))`.
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