Question

A body is projected such that it's kinetic energy at the top 3/4th of it's initial kinetic energy . What is the initial angle of projection of the projectile with the horizontal.

Answer 1

KE(initial) $=\frac{1}{2}m{u}^2$

KE(final) $=\frac{1}{2}m(u\cos \theta {)}^2$

$\frac{1}{2}m{u}^2{\cos }^2\theta =\frac{3}{4}\times \frac{1}{2}m{u}^2$

$\cos \theta =\frac{\sqrt{3}}{2}$

$\theta ={30}^{\circ }$