Q 11-02-101JEE MainJEE Main 2022 (28 Jul, Shift 1)Easy
A NCC parade is going at a uniform speed of $9\ \text{km h}^{-1}$ under a mango tree on which a monkey is sitting at a height of $19.6$ m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is :
(Given $g = 9.8\ \text{m s}^{-2}$)
Answer: (A) $5$ m
Time for the mango to fall:
$$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times19.6}{9.8}} = 2\ \text{s}$$
The parade moves at $9\ \text{km h}^{-1} = 2.5\ \text{m s}^{-1}$, so the cadet who catches it was $2.5\times2 = 5$ m from the tree when it was dropped.
Solution by Sreeraj P, M.Sc Physics