# Problematic Balloon

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.

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

3 years, 11 months ago

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Well 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.

- 3 years, 11 months ago

from which source did u get the problem

- 2 years ago

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 $$t$$ $$=$$ $$0$$, $$h$$ $$=$$ $$0$$

- 3 years, 11 months ago