Kinematics

JEE Main level, numerical answer, medium

171 of 308

Question 171 of 308

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6s6s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5s1.5s. A third ball released, from the rest from the same location, will reach the ground in _____ s.

Show answer and solution

Answer: 3

When it is thrown upward, −h=v×6−12g(6)2-h=v\times 6-\dfrac{1}{2}g{\left(6\right)}^{2}
When it is thrown downward, −h=−v×1.5−12g(1.5)2-h=-v\times 1.5-\dfrac{1}{2}g{\left(1.5\right)}^{2}
Therefore, 5h=5[36+9]5h=5\left[36+9\right] h=45mh=45m
When the ball is dropped, h=12gt2⇒45=12×10×t2⇒t=3sh=\dfrac{1}{2}g{t}^{2}\Rightarrow 45=\dfrac{1}{2}\times 10\times {t}^{2}\Rightarrow t=3s
Alternate method, t=t1t2=6×1.5=3st=\sqrt{{t}_{1}{t}_{2}}=\sqrt{6\times 1.5}=3s

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.