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(C) Let the height of the balloon when the stone is dropped be $h$ and its upward velocity be $v$. When the stone is dropped, its initial velocity is

Vedclass · Accepted Answer

$256$

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(C) Let the height of the balloon when the stone is dropped be $h$ and its upward velocity be $v$. When the stone is dropped, its initial velocity is $u = v$ (upwards). Using the equation of motion for the stone: $s = ut + \frac{1}{2}at^2$. Here, $s = -h$ (displacement is downwards), $u = v$, $a = -g = -32 \ ft/sec^2$, and $t = 4 \ sec$. $-h = v(4) + \frac{1}{2}(-32)(4)^2$. $-h = 4v - 16(16)$. $-h = 4v - 256$. $h = 256 - 4v$. In $4 \ sec$, the balloon rises further by a distance $d = v 	imes t = v 	imes 4 = 4v$. The total height of the balloon from the ground when the stone hits the ground is $H = h + d$. $H = (256 - 4v) + 4v = 256 \ ft$.

From a balloon rising vertically with a uniform velocity of $v \ ft/sec$, a stone is dropped. If the stone reaches the ground after $4 \ sec$, what is the height of the balloon above the ground at that moment (in $ft$)? (Take $g = 32 \ ft/sec^2$)

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