Kinematics

JEE Main level, single correct, medium

157 of 308

Question 157 of 308

An object is allowed to fall from a height RR above the earth, where RR is the radius of earth. Its velocity when it strikes the earth’s surface, ignoring air resistance, will be :

Show answer and solution

Answer: (B) gR\sqrt{gR}

As gravitational field is a conservative field and therefore mechanical energy will be conserved. Therefore,
Loss in PE = Gain in KE
(−GMm2R)−(−GMmR)=12mv2\left(-\dfrac{GMm}{2R}\right)-\left(-\dfrac{GMm}{R}\right)=\dfrac{1}{2}m{v}^{2}
⇒v2=GMR=gR(asg=GMR2)\Rightarrow {v}^{2}=\dfrac{GM}{R}=gR\left(\text{as}g=\dfrac{GM}{{R}^{2}}\right)
⇒v=gR\Rightarrow v=\sqrt{gR}

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.