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Physics · Motion in one Dimension

JEE Main 2026 — 8 April, Evening Shift — Question 4

A gas balloon is going up with a constant velocity of 10m/s10\mathrm{m/s}. When this balloon reached a height of 75m75\mathrm{m}, a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is m\mathrm{m}. (Take g=10m/s2\mathrm{g} = 10\mathrm{m/s}^2)

  1. Option A:

    85

  2. Option B:

    150

  3. Option C:

    129

  4. Option D:

    125

    Correct

Answer: D

Step-by-step solution

For stone: initial velocity upward 10 m/s, initial height 75 m, acceleration -10 m/s². Using s=ut+12at2s = ut + \frac12 at^2, −75=10t−5t2-75 = 10t -5t^2 ⇒ t2−2t−15=0t^2 -2t -15=0 ⇒ t=5t=5 s. Balloon rises additional 10×5=5010\times5=50 m, so height = 75+50=12575+50=125 m.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
8 April, Evening Shift
Subject
Physics
Chapter
Motion in one Dimension
Topic
Motion Under Gravity

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