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

Step 1 : we know that, velocity of a projectile at half of maximum height `u= sqrt((1+ cos^(2) theta)/(2))`
Step 2 : given that `u cos theta sqrt((2)/(5)) xx u sqrt(((1+ cos^(2) theta))/(2))`
Squaring on both sides `u^(2) cos^(2) theta=(2)/(5) u^(2)((1+cos^(2) theta)/(2))`
`10 cos theta=2+ 2 cos theta`
`rArr 8 cos^(2) theta=2 rArr cos^(2) theta =(1)/(4) rArr theta =60^(@)`