Kinematics

JEE Main level, single correct, medium

241 of 308

Question 241 of 308

Two projectile thrown at 30∘30^{\circ} and 45∘45^{\circ} with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

Show answer and solution

Answer: (C) 2:1\sqrt{2}:1

Time taken to reach maximum height is given by
t=usin⁡θgt=\dfrac{u\sin \theta }{g}, where, uu is initial velocity.
Here, t1=t2{t}_{1}={t}_{2}
∴u1sin⁡θ1g=u2sin⁡θ2g∴\dfrac{{u}_{1}\sin {\theta }_{1}}{g}=\dfrac{{u}_{2}\sin {\theta }_{2}}{g}
⇒u1sin⁡30=u2sin⁡45\Rightarrow {u}_{1}\sin 30={u}_{2}\sin 45
⇒u1u2=1212=21=2:1\Rightarrow \dfrac{{u}_{1}}{{u}_{2}}=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}}=\dfrac{\sqrt{2}}{1}=\sqrt{2}:1

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 (Projectile motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.