QJEEM22Ph8d8b: JEE Main 2022 Physics - Class 11 JEEM22Ph8d8b Solution: A juggler throws balls vertically upwards with same initial…

The juggler throws n balls per second.

$\therefore$ Interval between balls = ${1 \over n}$ seconds

At maximum height velocity of first ball $v = 0$

Juggler throws all balls with same initial velocity $= u$

For 1st ball,

${u_1} = u$ , ${v_1} = 0$ , $a = - g$ , $S = {H_{\max }}$

Using formula,

${v^2} = {u^2} + 2as$

$0 = {u^2} - 2g\,{H_{\max }}$

$\Rightarrow {H_{\max }} = {{{u^2}} \over {2g}}$ ...... (1)

And using formula,

$v = u + at$

$\Rightarrow 0 = u - gt$

$\Rightarrow t = {u \over g}$

$\therefore$ Time taken by ball 1 to reach maximum height $({H_{\max }}) = {u \over g}$

According to the question,

$t = {1 \over n}$

$\Rightarrow {u \over g} = {1 \over n}$

$\Rightarrow u = {g \over n}$ ....... (2)

Putting value of $u$ in equation (1), we get

${H_{\max }} = {{{{\left( {{g \over n}} \right)}^2}} \over {2g}}$

$= {{{g^2}} \over {2{n^2}g}}$

$= {g \over {2{n^2}}}$