Q 11-02-044JEE MainJEE Main 2026 (8 Apr, Shift 2)Medium
A gas balloon is going up with a constant velocity of $10\ \text{m/s}$. 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 $g = 10\ \text{m/s}^2$)
Answer: (D) $125$
When released, the stone has the balloon's velocity: $10$ m/s upward. Taking upward as positive, it reaches the ground ($-75$ m) when
$$-75 = 10t - 5t^2 \;\Rightarrow\; t^2 - 2t - 15 = 0 \;\Rightarrow\; t = 5\ \text{s}$$
In these $5$ s the balloon rises another $10 \times 5 = 50$ m, so its height is $75 + 50 = 125$ m.
Solution by Sreeraj P, M.Sc Physics