JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Kinematics

An object is projected with kinetic energy K from a point A at an angle

6 0^{\circ}

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

A

the kinetic energy is

K = \frac{1}{2} m u^{2}

. At the top

B

only the horizontal component of velocity survives,

u_{B} = u cos 6 0^{\circ} = \frac{u}{2}

, so

(K E)_{B} = \frac{1}{2} m (\frac{u}{2})^{2} = \frac{K}{4}

. The point

C

is at the same height as

A

, so by energy conservation

(K E)_{C} = K

. The difference is

(K E)_{C} - (K E)_{B} = K - \frac{K}{4} = \frac{3 K}{4}

, and its ratio to the kinetic energy at

A

\frac{3 K /4}{K} = \frac{3}{4}

, that is

3 : 4

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option

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