Question

It's just a question from projectile trajectory.. Could anyone help me out with this problem please :'(

Answer 1

Initial kinetic energy = E = $\frac{1}{2} mu^{2}$

At the highest point the body will only have horizontal component of velocity
$v = ucos\:30^{o} = \frac{ \sqrt{3}u }{2}$

Kinetic Energy at highest point = $\frac{1}{2} m( \frac{ \sqrt{3} u}{2} )^{2} = \frac{1}{2} mu^{2} \times \frac{3}{4} = \frac{3E}{4}$