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The horizontal range and the maximum hei...

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

A

`theta = tan^(-1)(1)`

B

`theta = tan^(-1)(2)`

C

`theta = tan^(-1)(3)`

D

`theta = tan^(-1)(4)`

Text Solution

Verified by Experts

The correct Answer is:
D
Horizontal rane `=(u^(2) sin 2theta)/R`
Maximum height `=(u^(2) sin 2 theta)/ g`
Both are equal (given)
`therefore (u^(2) sin 2 theta)/g = (u^(2) sin^(2) theta)/(2g) rArr 2sin theta cos theta =(sin^(2) theta)/2`
`therefore tan theta = 4 rArr theta = tan^(-1)(4)`
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