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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 ((2u^2 )/g) horizontally  away from it. Find the height of the  balloon as a function of time.

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

Answered by mrW1 last updated on 19/Jul/17

v=(dy/dt)  T=time the stone needs to hit the ground  uT=((2u^2 )/g) as given  ⇒T=((2u)/g)  y=−vT+(1/2)gT^2   ⇒y=−((2u)/g)×(dy/dt)+(1/2)g×((4u^2 )/g^2 )=−((2u)/g)((dy/dt)−u)  ⇒(dy/dt)=u−(g/(2u))y  (dy/(u−(g/(2u))y))=dt  −((2u)/g)ln (u−(g/(2u))y)=t+C  y=0 at t=0  ⇒C=−((2u)/g)ln u  ((2u)/g)[ln u−ln (u−(g/(2u))y)]=t  ((2u)/g)[ln (u/(u−(g/(2u))y))]=t  (u/(u−(g/(2u))y))=e^((g/(2u))t)   u−(g/(2u))y=ue^(−(g/(2u))t)   ⇒y=((2u^2 )/g)(1−e^(−((gt)/(2u))) )

$$\mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}} \\ $$$$\mathrm{T}=\mathrm{time}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{needs}\:\mathrm{to}\:\mathrm{hit}\:\mathrm{the}\:\mathrm{ground} \\ $$$$\mathrm{uT}=\frac{\mathrm{2u}^{\mathrm{2}} }{\mathrm{g}}\:\mathrm{as}\:\mathrm{given} \\ $$$$\Rightarrow\mathrm{T}=\frac{\mathrm{2u}}{\mathrm{g}} \\ $$$$\mathrm{y}=−\mathrm{vT}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gT}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{y}=−\frac{\mathrm{2u}}{\mathrm{g}}×\frac{\mathrm{dy}}{\mathrm{dt}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}×\frac{\mathrm{4u}^{\mathrm{2}} }{\mathrm{g}^{\mathrm{2}} }=−\frac{\mathrm{2u}}{\mathrm{g}}\left(\frac{\mathrm{dy}}{\mathrm{dt}}−\mathrm{u}\right) \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{u}−\frac{\mathrm{g}}{\mathrm{2u}}\mathrm{y} \\ $$$$\frac{\mathrm{dy}}{\mathrm{u}−\frac{\mathrm{g}}{\mathrm{2u}}\mathrm{y}}=\mathrm{dt} \\ $$$$−\frac{\mathrm{2u}}{\mathrm{g}}\mathrm{ln}\:\left(\mathrm{u}−\frac{\mathrm{g}}{\mathrm{2u}}\mathrm{y}\right)=\mathrm{t}+\mathrm{C} \\ $$$$\mathrm{y}=\mathrm{0}\:\mathrm{at}\:\mathrm{t}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{C}=−\frac{\mathrm{2u}}{\mathrm{g}}\mathrm{ln}\:\mathrm{u} \\ $$$$\frac{\mathrm{2u}}{\mathrm{g}}\left[\mathrm{ln}\:\mathrm{u}−\mathrm{ln}\:\left(\mathrm{u}−\frac{\mathrm{g}}{\mathrm{2u}}\mathrm{y}\right)\right]=\mathrm{t} \\ $$$$\frac{\mathrm{2u}}{\mathrm{g}}\left[\mathrm{ln}\:\frac{\mathrm{u}}{\mathrm{u}−\frac{\mathrm{g}}{\mathrm{2u}}\mathrm{y}}\right]=\mathrm{t} \\ $$$$\frac{\mathrm{u}}{\mathrm{u}−\frac{\mathrm{g}}{\mathrm{2u}}\mathrm{y}}=\mathrm{e}^{\frac{\mathrm{g}}{\mathrm{2u}}\mathrm{t}} \\ $$$$\mathrm{u}−\frac{\mathrm{g}}{\mathrm{2u}}\mathrm{y}=\mathrm{ue}^{−\frac{\mathrm{g}}{\mathrm{2u}}\mathrm{t}} \\ $$$$\Rightarrow\mathrm{y}=\frac{\mathrm{2u}^{\mathrm{2}} }{\mathrm{g}}\left(\mathrm{1}−\mathrm{e}^{−\frac{\mathrm{gt}}{\mathrm{2u}}} \right) \\ $$

Commented by Tinkutara last updated on 19/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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