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 / s² )

Solution

Physics sir · 4 steps

The mango's fall time is fixed by height alone. The cadet just has to walk into place during that time.

1. Time to fall. The mango is dropped from rest, so it has no horizontal velocity. Its fall time depends only on the height: h = 1/2gt².

   t = √(2h/g) = √((2 × 19.6)/9.8) = √4 = 2 s

2. Convert the speed. 9 km/h = 9 × 5/18 = 2.5 m s⁻¹.
3. Distance covered. In 2 s the cadet walks 2.5 × 2 = 5 m. To be under the tree when the mango lands, he must be 5 m away when it is dropped.
4. So. The cadet who receives the mango is 5 m from the tree at the moment of the drop, option (a).

The idea to keep. In catch-the-falling-object questions, find the fall time from the height first. The catcher's distance is then just speed times that time. Convert km/h to m/s with 5/18.

Answer from the NTA final answer key. Solution by PrepWiser.