A particle is projected at an angle `theta` with the horizontal . If angle of elevation of highest point of trajectory is `phi` when seen from point of projection, then

Options

A. `tan phi =2 tan phi`
B. `tan theta =3 tan phi`
C. `tan theta =2 tan phi`
D. `tan theta=4 tan phi`

Answer

C

Explanation

(c): Let particle be projected with a speed u

From the figure,
`tan phi =(H)/(R//2)=(2H)/(R)=((2(u^(2) sin^(2)theta)/(2g))/(u^(2) sin 2 theta))/(g)`
`tan phi =(sin^(2) theta)/(sin 2theta)=(sin ^(2)theta)/(2 sin theta cos theta)=(1)/(2)tan theta`
Hence, option (c) is correct .