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In a circus, a performer throws an apple towards a hoop held at $45 \,m$ height by another performer standing on a high platform (see figure). The thrower aims for the hoop and throws the apple with a speed of $24 \,m / s$. At the exact moment that the thrower releases the apple, the other performer drops the hoop. The hoop falls straight down. At ............ $m$ height above the ground does the apple go through the hoop?
$21$
$22$
$23$
$24$
Solution
(b)
Given, velocity of projection of apple, $v=24 \,m / s$
Distance between point of projection and hoop is given by
$A C=\sqrt{(25)^2+(45)^2}=\sqrt{2650} m$
So, time taken by the ball to the hoop,
$t=\frac{\sqrt{2650}}{24}$ $s$
The distance covered by hoop is
$s=\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times(\frac{\sqrt{2650}}{{24}})^2$
$=5 \times \frac{2650}{576}=\frac{13250}{576}$
$\therefore$ Height above the ground where apple go through the hoop is
$H=45-\frac{13250}{576}$
$=\frac {12670}{576}\simeq 22 \,m$
