JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Kinematics

A NCC parade is going at a uniform speed of

9 km / h

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}

)

(A)5 m✓ Correct
(B)10 m
(C)19.8 m
(D)24.5 m

Answer: (A)

Solution

The mango falls freely from height

h = 19.6

m. Time of fall:

t = \frac{2 h}{g} = \frac{2 \times 19.6}{9.8} = 4 = 2

s. The parade moves horizontally at

v = 9

km/h

= 9 \times \frac{5}{18} = 2.5

m/s. Distance covered in 2 s:

d = v t = 2.5 \times 2 = 5

m. The cadet who is 5 m from the tree at the moment of drop will receive the mango.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option

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