A ball having kinetic energy KE, is projected at an angle of 60° from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?

Solution

Physics sir · 3 steps

At the top only the horizontal velocity survives, and cos 60° = 1/2, so KE drops to a quarter.

1. At the top. Gravity acts only vertically, so the horizontal velocity u cos 60° never changes. At the highest point the vertical velocity is zero, so the speed there is just u cos 60° = u/2.

   KEₜₒₚ = 1/2m(u cos 60°)² = KE cos² 60°

2. Put in numbers. cos² 60° = (1/2)² = 1/4. Kinetic energy goes as speed squared, so halving the speed cuts KE to one quarter.
3. So. The kinetic energy at the highest point is KE/4, option (c).

The idea to keep. At the highest point, speed is u cos θ, so kinetic energy is KE cos² θ. Always square the cosine, because KE depends on v².

Answer from the NTA final answer key. Solution by PrepWiser.