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 points B and C to that at point A (see figure), in the absence of air friction is :
(A)1:2
(B)2:3
(C)1:4
(D)3:4✓ Correct
Answer: (D)
Solution
At A the kinetic energy is K=21mu2. At the top B only the horizontal component of velocity survives, uB=ucos60∘=2u, so (KE)B=21m(2u)2=4K. The point C is at the same height as A, so by energy conservation (KE)C=K. The difference is (KE)C−(KE)B=K−4K=43K, and its ratio to the kinetic energy at A is K3K/4=43, that is 3:4.
JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option