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The horizontal range of a projectile is ...

The horizontal range of a projectile is `4 sqrt(3)` times its maximum height. Its angle of projection will be

A

`45^(@)`

B

`60^(@)`

C

`90^(@)`

D

`30^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
`R = 4sqrt(3) H`
`rArr (u^(2) sin 2theta)/g = 4sqrt(3) (u^(2) sin^(2) theta)/(2g)`
`rArr tan theta = 1/sqrt(3) rArr theta = 30^(@)`
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