JEE Main 2025 Jan 23 Shift 2, Physics Q1: Motion In Two And Three Dimension

JEE Main2025Jan 23, Shift 2Physics

A ball having kinetic energy KE, is projected at an angle of $6 0^{\circ}$ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?

A
$\frac{( K E )}{8}$
B
$\frac{( K E )}{4}$
C
$\frac{( K E )}{16}$
D
$\frac{( K E )}{2}$

Solution

At the highest point of a projectile, only the horizontal component of velocity survives.

If launch speed is $u$ , then $v_{t o p} = u cos 6 0^{\circ} = \frac{u}{2}$ .

$K E_{t o p} = \frac{1}{2} m v_{t o p}^{2} = \frac{1}{2} m (\frac{u}{2})^{2} = \frac{1}{4} \cdot \frac{1}{2} m u^{2} = \frac{K E}{4} .$

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JEE Main 2025 Jan 23 Shift 2, Physics Q1: Motion In Two And Three Dimension

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