Question

A projectile is projected with kinetic energy K. If it attains maximum range, then what will be its kinetic energy at maximum height?

Answer 1

If the velocity of projection be v and mass of the projectile be m, then \(K = \frac{1}{2}mv^2\)

∵ The projectile attains maximum range, therefore angle of projection will be θ = 45°.

∴ At the time of projection,

\(v_x = v\cos45^\circ = \frac{v}{\sqrt{2}}\) (horizontal velocity)

and v_y = v sin 45° = \(\frac{v}{\sqrt{2}}\) (vertical velocity)

∵ The horizontal velocity v_x remains unchanged and at heighest point, v_y becomes zero. Only v_x remains there.

∴ Kinetic energy at highest point,

\(K' = \frac{1}{2}mv_x{}^2 = \frac{1}{2}m(\frac{v}{\sqrt{2}})^2 = \frac{1}{2}\ \frac{mv^2}{2}\)

or \(K' =\frac{K}{2}\) 

or K' = 0.5 K