Question 42: Motion in 1D

Question

A NCC parade is going at a uniform speed of $$9 \mathrm{~km} / \mathrm{h}$$ under a mango tree on which a monkey is sitting at a height of $$19.6 \mathrm{~m}$$. 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 $$g=9.8 \mathrm{~m} / \mathrm{s}^{2}$$ )

Accepted Answer

A NCC parade is moving at a steady speed of 9 km/h under a mango tree where a monkey is perched at a height of 19.6 meters. Suddenly, the monkey drops a mango. To determine which cadet will catch the mango, we need to calculate the distance of the cadet from the tree at the moment the mango is dropped, considering the acceleration due to gravity $ g = 9.8 \, 	ext{m/s}^2 $.
First, we use the formula to find the time $ t $ it takes for the mango to fall:
$ H = \frac{1}{2}\,g\,t^2 $
Substituting the given values:
$ 19.6 = 4.9\,t^2 $
Solving for $ t $:
$ t^2 = \frac{19.6}{4.9} $
$ t^2 = 4 $
$ t = 2 \, 	ext{sec} $
Next, we need to find the distance $ D $ the cadet travels in these 2 seconds:
$ D = 	ext{speed} 	imes 	ext{time} $
Since the speed is given in km/h, we first convert it to m/s:
$ 9 \, 	ext{km/h} = 9 	imes \frac{1000}{3600} \, 	ext{m/s} = 2.5 \, 	ext{m/s} $
So, the distance covered by the cadet is:
$ D = 2.5 \, 	ext{m/s} 	imes 2 \, 	ext{sec} = 5 \, 	ext{m} $
Thus, the cadet who is 5 meters away from the tree at the moment the mango is dropped will catch the mango.

Answer

5 m

Answer

10 m

Answer

19.8 m

Answer

24.5 m

Motion in 1D

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