JEE Main 2026 Apr 8 Shift 2, Physics Q4: Motion In One Dimension

JEE Main2026Apr 8, Shift 2Physics

A gas balloon is going up with a constant velocity of $10$ m/s. When this balloon reached a height of $75$ 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. (Take $g = 10$ m/s $^{2}$ )

A
$85$
B
$150$
C
$129$
D
$125$

Solution

At the moment of release the stone shares the balloon's upward velocity of $+ 10$ m/s. Taking upward as positive and the release point as origin, the ground is at displacement $- 75$ m.

$- 75 = 10 t - \frac{1}{2} (10) t^{2}$

$t^{2} - 2 t - 15 = 0 ⟹ (t - 5) (t + 3) = 0 ⟹ t = 5$ s.

During this $5$ s the balloon continues at $10$ m/s, climbing an additional $10 \times 5 = 50$ m above its release point ( $75$ m).

$∴$ Height of balloon when stone lands $= 75 + 50 = 125$ m.

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JEE Main 2026 Apr 8 Shift 2, Physics Q4: Motion In One Dimension

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