Question

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 below). The thrower aim $s$ for the hoop and throws the apple with a speed of $24\, m / s$. At the exact moment that the thrower released the apple, the other performer drops the hoop. The hoop falls straight down. At what height above the ground does the apple go through the hoop ?

• A. 21 m
• B. 22 m
• C. 23 m
• D. 24 m

Answer 1

Answer: (B)

Velocity of projection $=24\, m / s$
Distance between point of projection and hoop
$=\sqrt{25^{2}+45^{2}}$
$\therefore $ Time taken by ball to reach the hoop
$=\frac{\sqrt{25^{2}+45^{2}}}{24}$
(Note :- We are analysing the motion wrt hoop)
$\therefore $ Distance by which hoop will fall
$=\frac{1}{2} at ^{2}=\frac{1}{2} \times 10 \times \frac{\left(25^{2}+45^{2}\right)}{24^{2}}$
$\therefore $ Height above the ground where apple go through the hoop is given by
$45-\left[\frac{1}{2} \times 10 \times \frac{\left(25^{2}+45^{2}\right)}{24^{2}}\right]=22\, m$