A gas balloon is going up with a constant vel | vedclass

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(D) Let the upward direction be positive. The initial velocity of the stone is $u = 10 	ext{ m/s}$ (same as the balloon).The displacement of the ston

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(D) Let the upward direction be positive. The initial velocity of the stone is $u = 10 	ext{ m/s}$ (same as the balloon).The displacement of the stone when it hits the ground is $s = -75 	ext{ m}$.Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $a = -g = -10 	ext{ m/s}^2$:$-75 = 10t - 5t^2$$5t^2 - 10t - 75 = 0$$t^2 - 2t - 15 = 0$$(t - 5)(t + 3) = 0$Since time cannot be negative,$t = 5 	ext{ s}$.In this time,the balloon continues to move upward at a constant velocity of $10 	ext{ m/s}$.The additional height gained by the balloon is $h = v 	imes t = 10 	ext{ m/s} 	imes 5 	ext{ s} = 50 	ext{ m}$.The total height of the balloon when the stone hits the ground is $75 	ext{ m} + 50 	ext{ m} = 125 	ext{ m}$.

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