A ball whose kinetic energy is E , is thrown at an angle of 45 ° with the horizontal. Its kinetic energy at the highest point of its flight will be:

Step-by-step solution

Kinetic energy is E 1 2 m v 2 = E ⇒ v = 2 E m Velocity at highest point = v cos 45 ° ∴ v cos 45 ° = 2 E m . 1 2 = E m [ ∵ at the highest point there is only horizontal velocity] ∴ K.E. at highest point = 1 2 × m E m = E 2