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

Explanation

Step 1: we know that, velocity of a projectile at half of maximum height: `=usqrt((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 =2/5u^(2)(1+cos^(2)theta)/2`
`10cos^(2)theta =2 + 2 cos^(2)theta rArr theta=60^(@)`