A balloon moves up vertically such that if a stone is projected with a horizontal velocity u relative to balloon, the stone always hits the ground at a fixed point at a distance ((2u^2 )/g) horizontally away from it. Find the height of the balloon as a function of time.


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Question Number 18274 by Tinkutara last updated on 17/Jul/17

$A balloon moves up vertically such$ $that if a stone is projected with a$ $horizontal velocity u relative to balloon,$ $the stone always hits the ground at a$ $fixed point at a distance \frac{2 u^{2}}{g} horizontally$ $away from it . Find the height of the$ $balloon as a function of time .$
	Answered by mrW1 last updated on 19/Jul/17

	$v = \frac{dy}{dt}$ $T = time the stone needs to hit the ground$ $uT = \frac{{2 u}^{2}}{g} as given$ $\Rightarrow T = \frac{2 u}{g}$ $y = - vT + \frac{1}{2} {gT}^{2}$ $\Rightarrow y = - \frac{2 u}{g} \times \frac{dy}{dt} + \frac{1}{2} g \times \frac{{4 u}^{2}}{g^{2}} = - \frac{2 u}{g} (\frac{dy}{dt} - u)$ $\Rightarrow \frac{dy}{dt} = u - \frac{g}{2 u} y$ $\frac{dy}{u - \frac{g}{2 u} y} = dt$ $- \frac{2 u}{g} \ln (u - \frac{g}{2 u} y) = t + C$ $y = 0 at t = 0$ $\Rightarrow C = - \frac{2 u}{g} \ln u$ $\frac{2 u}{g} [\ln u - \ln (u - \frac{g}{2 u} y)] = t$ $\frac{2 u}{g} [\ln \frac{u}{u - \frac{g}{2 u} y}] = t$ $\frac{u}{u - \frac{g}{2 u} y} = e^{\frac{g}{2 u} t}$ $u - \frac{g}{2 u} y = {ue}^{- \frac{g}{2 u} t}$ $\Rightarrow y = \frac{{2 u}^{2}}{g} (1 - e^{- \frac{gt}{2 u}})$
	Commented by Tinkutara last updated on 19/Jul/17

	$Thanks Sir!$
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