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A projectile is fired at an angle of 45^...

A projectile is fired at an angle of `45^(@)` with the horizontal. Elevation angle of the projection at its highest point as seen from the point of projection is

A

`45^(@)`

B

`60^(@)`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C

Let `phi` be elevation ange ofthe projec tile at its highest point as secn from the point of projection O and `theta` be angle of projection with the horizontal
`tan phi=(H)/(R//2)`
In case of projectile motion
Maximum heights, `H=(u^(2)sin^(2)theta)/(g)`
Horizontal range, `R(u^(2) sin^(2) 2theta)/(g)`
Substituting these value of H and R in eqm., we get,
`tan phi=((u^(2) sin theta)/(2g))/((u^(2)sin 2 theta)/(2g))`
`tan phi=(sin^(2) theta)/(sin 2theta)=(sin^(2)theta)/(2 sin theta cos theta) =(1)/(2) tan theta`
Here `theta=45^(@)`
`tan phi=(1)/(2) tan 45^(@)=(1)/(2)`
`:. tan theta 45^(@)=1`
`phi= tan^(-1)((1)/(2))`
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