×

# 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, 6 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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, 6 months ago

from which source did u get the problem

- 1 year, 7 months 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, 6 months ago