A NCC parade is going at a uniform speed of $9 km h^{- 1}$ 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 $g = 9.8 m s^{- 2}$ )

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TestHub · Accepted Answer

If a cadet receives the mango, then the time taken by the mango to fall and the cadet to reach there should be equal.
Time taken by mango to fall $t = \sqrt{\frac{2 h}{g}}$ .
$\Rightarrow t = \sqrt{\frac{2 \times 19.6}{9.8}} = 2 s$
Therefore, the distance of the cadet from the tree will be, $d = v t$ .
$\Rightarrow d = 9 \times \frac{5}{18} \times 2 = 5 m$

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