Question

A juggler keeps $5$ balls going with one hand , so that at any instant, $4$ball in the hand. If each ball rises to a height of $20$m, then the time for which each ball stays in the hand is $(g=10m/s^2)$

• A. $1.0s$
• B. $1.5s$
• C. $2.0s$
• D. $4.0s$

Answer 1

The correct option is A $1.0s$

Let ${}^{\prime }{n}^{\prime }$ be the initial velocity of the ball while going upwards

The final velocity of ball at height in $20$ $V+0$

$V^2=u^2+2as$

$O=u^2+2as$

$u=\sqrt{2gx}=g$

$=\sqrt{2\times 10\times 20}$

$u=20$

If ${}^{\prime }{t}^{\prime }$ in taken be the ball going up through distance $20mO=u+(-g)t$ or $t=u/g=\frac{20}{10}=2\sec$

$=2\sec$

Total time after which the ball comes into hand

$T=2t=\frac{2u}{g}=\frac{2}{g}\sqrt{2g}20=2\frac{2\sqrt{2\times 20}}{g}=\frac{2\sqrt{2\times 20}}{g}$

$=2\times 2$

$=4\sec$

During $T_{1}4$ ball will be in air and one ball in hand for one ball in hand

$\frac{T}{4}=\frac{4}{4}=1\sec$

Time for $4$ balls $4\sec$ one ball $1\sec$