Help me with this one.

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

The answer is

\(\frac{2u^2}{g}\)(\(1-e^{-gt/2u}\))

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## Comments

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TopNewestWell this is not very difficult.

Let the velocity of the balloon when it is at a height \(h\) be \(v\). Hence when the stone is thrown from the balloon with a velocity of \(u\) in the horizontal direction, relative to the balloon, the actual velocity of the stone, with respect to the ground will be \(u\) in the horizontal direction and \(v\) in the vertical. Now, they say that irrespective of the height of the balloon above the ground the range of the stone is \(\displaystyle \frac{2u^2}{g} \). Hence if the balloon takes a time \(t_0 \) to reach the ground then we can easily write \(\displaystyle u t_0 = \frac{2u^2}{g} \Rightarrow t_0 = \frac{2u}{g} \). Also using the second equation of motion in the vertical direction, we obtain \( \displaystyle h = -v t_0 + \frac{1}{2} gt_0^2 \). Substitute the value of \(t_0\) to obtain an equation of the form

\(\displaystyle \Rightarrow v + \frac{gh}{2u} = u \)

\(\displaystyle \Rightarrow \frac{\text{d}h}{\text{d}t} + \frac{gh}{2u} = u \)

This is a standard differential equation with the solution of the form \( \displaystyle h.\text{exp}(\frac{gt}{2u}) = ut + C \) , where \(C\) is the constant of integration. Rearrange to obtain the final answer as \[ h = (ut + C) \text{exp}(\frac{-gt}{2u}) \]

(Note: \( \text{exp}(x) = e^x \) )

Hope this helps.

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from which source did u get the problem

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I got a different differential equation than yours. It is

\(h\) \(=\) \(\frac{2u^2}{g}\) + \(c\) \(e^{(-gt/2u)}\)

I am just having problem in finding the constraint. I am confused whether to take

when \(t\) \(=\) \(t_{0}\), \(h\) \(=\) \(0\)

or, when \(t\) \(=\) \(0\), \(h\) \(=\) \(0\)

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