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-horizontal

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!