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 :

Step-by-step solution

At A: KE_A = K , v_A = v , angle = 60° . At B (highest point): v_B = v 60° = v/2 , so KE_B = K/4 . At C (same level as A): by energy conservation, KE_C = K . |KE_B - KE_C| = |K/4 - K| = 3K/4 . Ratio = 3K/4 K = 3 4 , i.e., 3:4 .