The total speed of a projectile at its greater height is `sqrt(6/7)` of its speed when it is at half of its greatest height. The angle of projection will be :-

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `50^(@)`

Answer

C

Explanation

`v_(y)^(2)=u^(2)sin^(2)theta-2gxxH/2,v_(x)^(2)=u^(2) cos^(2) theta`

`:. U cos theta =sqrt(6/7)[sqrt(v_(x)^(2)+v_(y)^(2))]rArr cos theta=sqrt(3)/2` or `theta=30^(@)`