An object is projected with kinetic energy K from a point A at an angle 60° with the horizontal. The ratio of the difference in kinetic energies at points B and C to that at point A (see figure), in the absence of air friction is :

Solution

Physics sir · 4 steps

At the top only the horizontal velocity survives, so the KE there is K times cos squared 60 degrees.

1. Speed at C. With no air friction, mechanical energy is conserved. C is at the same height as A, so the speed at C equals the speed at A and K_C = K.
2. Speed at B. At the highest point the vertical velocity is zero. Only the horizontal part u cos 60° remains, and it never changes because gravity acts vertically.

   K_B = 1/2m(u cos 60°)² = K cos² 60° = K/4

3. Take the difference. K_C - K_B = K - K/4 = 3K/4. Compare this with K_A = K.
4. So. The ratio is (3K/4)/K = 3:4, which is option (a).

The idea to keep. For a projectile, KE at the top is K cos² θ because only the horizontal velocity is left. KE at any point on the launch level is the same as at launch.

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