Question

An 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: 
Give g=9.8 m/s²

• A. 5 m
• B. 10 m
• C. 19.8 m
• D. 24.5 m

Answer 1

The Correct Option is A

To solve this problem, we must determine the distance from the tree where a cadet should be to catch a mango dropped from a height, given the uniform speed of the parade and the time the mango takes to fall.

1. Given:
   • Height of the mango on the tree, \(h = 19.6 \, \text{m}\)
   • Acceleration due to gravity, \(g = 9.8 \, \text{m/s}^2\)
   • Speed of the parade, \(v = 9 \, \text{km/h} = \frac{9 \times 1000}{60 \times 60} \, \text{m/s} = 2.5 \, \text{m/s}\)
2. Using the second equation of motion to find the time \((t)\) it takes for the mango to fall:

\(h = \frac{1}{2}gt^2 \implies 19.6 = \frac{1}{2} \times 9.8 \times t^2 \implies 19.6 = 4.9t^2\)

\(t^2 = \frac{19.6}{4.9} = 4 \implies t = 2 \, \text{s}\)

1. Now, calculate the distance \((d)\) covered by a cadet in the time \(t = 2 \, \text{s}\):

\(d = v \times t = 2.5 \times 2 = 5 \, \text{m}\)

1. Conclusion: A cadet who is 5 meters away from the tree at the time the mango is dropped will be in the position to catch it.
2. Thus, the correct option is 5 m.

This solution helps us understand the physics of free fall and uniformly moving bodies, which is fundamental in solving many similar problems.