JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Kinematics

A ball is projected with kinetic energy E, at an angle of

6 0^{\circ}

to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :

(A)Zero
(B) $\frac{E}{2}$
(C) $\frac{E}{4}$ ✓ Correct
(D)E

Answer: (C)

Solution

Initial kinetic energy

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

. At the highest point of projectile motion, only the horizontal component of velocity remains:

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

. The kinetic energy at the highest point is

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

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option

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