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 ?

Options

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

Answer

B

Explanation

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
`=(sqrt(25^2+45^2))/24`
(Note :- We are analysing the motion wrt hoop)
`therefore` Distance by which hoop will fall
`=1/2at^2=1/2xx10xx((25^2+45^2))/(24^2)`
`therefore` Height above the ground where apple go through the hoop is given by
`45-[1/2xx10xx((25^2+45^2))/(24^2)]=22m`