The velocity of a projectile when it is at the greatest height is `(sqrt (2//5))` times its velocity when it is at half of its greatest height. Determine its angle of projection.

Explanation

`u=cos theta=sqrt(2/5)xxusqrt((1+cos^(2)theta)/2)`
Squaring on both sides
`u^(2) cos^(2) theta=2/5u^(2)((1+cos^(2) theta)/2)`
`10cos^(2) theta=2+2cos^(2) theta`
`rArr 8cos^(2)theta=2 rArr cos^(2) theta=1/4 rArr theta=60^(@)`