Kinematics

JEE Main level, single correct, medium

163 of 308

Question 163 of 308

A NCC parade is going at a uniform speed of 9kmh−19km{h}^{-1} under a mango tree on which a monkey is sitting at a height of 19.6m19.6m. 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=9.8ms−2g=9.8m{s}^{-2})

Show answer and solution

Answer: (A) 5m5m

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=2hgt=\sqrt{\dfrac{2h}{g}}.
⇒t=2×19.69.8=2s\Rightarrow t=\sqrt{\dfrac{2\times 19.6}{9.8}}=2s
Therefore, the distance of the cadet from the tree will be, d=vtd=vt.
⇒d=9×518×2=5m\Rightarrow d=9\times \dfrac{5}{18}\times 2=5m

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Projectile motion simulationDrag the sliders and watch the numbers in this question change.Revise Kinematics & Projectile Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Kinematics (Motion Under Gravity), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.