JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Kinematics

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✓ Correct

Answer: (D)

Solution

At the instant of release the stone shares the balloon's upward velocity

u = 10 m/s

and is at height

75 m

. Taking upward as positive, it reaches the ground when

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

, i.e.

t^{2} - 2 t - 15 = 0

, giving

t = 5 s

. During these

5 s

the balloon keeps rising at

10 m/s

and climbs a further

10 \times 5 = 50 m

, reaching

75 + 50 m

. The height of the balloon when the stone lands is

125 m

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option

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