Question

The velocity of a projectile when it is at half of the maximum height is 5/2 times it
velocity when it is at the heighest point. The angle of projection is​

Answer 1

Answer:

Time taken to reach max height =gusinθ Max  height =2gu2sin2θ

Half the max height =4gu2sin2θ Horizontal 'v' here ucosθ

Vertical velocity at half max height

=u2sin2θ−2g(2gusinθ)=2usinθ

Velocity at half the max-height

=u2cos2θ+2u2sin2θ⇒ucosθ=52u2cos2θ+2u2sin2θ⇒u2cos2θ=52u2(1−sin2θ