A gas balloon is going up with a constant velocity of $10 ext{ }m/s$. When this balloon reached a h

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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 $g = 10\text{ }m/s^{2}$ )

Accepted Answer

Sol. [Key Concept: Take the drop point as origin: $- 75 = 10t - \frac{1}{2}\ g\ t^{2}$ gives $t = 5$ s, so the balloon (still rising at $10$ m/s) is at $75 + 10 	imes 5 = 125$ m.]$${- 75 = 10 	imes t - \frac{1}{2} 	imes 10 	imes t^{2} }{t^{2} - 2t - 15 = 0 }{(t - 5)(t + 3) = 0 \Rightarrow t = 5 }$$Height $= 75 + 10 	imes 5 = 125	ext{ }m$

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