Question

A projectile leaves the ground at an angle 30° to the horizontal. The kinetic energy is E neglecting the air resistance . Find in terms of E it's kinetic energy at the highest point of motion

Answer 1

At highest point only vertical components of velocity

$v_x=vcos\theta=vcos30°=\frac{\sqrt3}{2}$

$E_{up}=\frac{1}{2}mv_x^2\\E_{up}=\frac{1}{2}mv^2cos^2\theta$

$E_{up}=\frac{1}{2}mv^2\times(\frac{\sqrt{3}}{2})^2=\frac{3}{8}mv^2$

$E=\frac{1}{2}mv^2$

$\frac{E_{up}}{E}=\frac{\frac{3}{8}mv^2}{\frac{mv^2}{2}}$

$E_{up}=\frac{3}{4}E$