JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Kinematics

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A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is

(A) $g /2 n$
(B) $g / n$
(C)2 gn
(D) $g /2 n^{2}$ ✓ Correct

Answer: (D)

Solution

The time taken by a ball to reach its highest point is

t = \frac{u}{g}

, where

u

is the initial velocity.

Since the juggler throws the next ball when the first reaches maximum height, the frequency of throwing is:

n = \frac{1}{t} = \frac{g}{u}

Therefore:

u = \frac{g}{n}

The maximum height reached is:

H_{max} = \frac{u ^{2}}{2 g} = \frac{( g / n ) ^{2}}{2 g} = \frac{g ^{2}}{2 g n ^{2}} = \frac{g}{2 n ^{2}}

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option

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