Question

the potential energy of a projectile at its highest point is 3 by 4 the value of its initial kinetic energy then the angle of projection​

Answer 1

At the highest point of the trajectory, the height of the projectile is,

H = u² sin²θ / 2g

Then the potential energy at the highest point is,

U = m g H

U = m g u² sin²θ / 2g

U = m u² sin²θ / 2

And the initial kinetic energy is,

K = m u² / 2

Given that,

U = 3 K / 4

m u² sin²θ / 2 = 3 m u² / 8

sin²θ = 3 / 4

Since θ is an acute angle,

sin θ = √3 / 2

This implies,

θ = 60°