A ballon moves up vertically such that if a stone is thrown from it with a horizontal velocity `v_(0)` relative to it the stone always hits the ground at a fixed point `2v_(0)^(2)//g` horizontally away from it. Find the height of the balloon as a function of time.

Explanation

Range `=v_(0)xx"time of flight" (t)`
`rArrt=(2v_(0)^(2))/(gv_(0))=(2v_(0))/(g)`
If `y` is the height of balloon at any instant `t` and `(dy)/(dt)` its velocity then
`y= -((dy)/(dt))((2v_(0))/(g))+(1)/(2)g((2v_(0))/(g))^(2)`
` rArr (dy)/(dt)+(g)/(2v_(0))y-v_(0)=0`
`(dy)/(v_(0)-(g)/(2v_(0))y)=dt`
` rArr -(2v_(0))/(g)ln(v_(0)-(g)/(2v_(0))y)=t+c`
At `t=0`, `y=0rArrc= -(2v_(0))/(g)lnv_(0)`
Simplfying we get
`y=(2v_(0)^(2))/(g)[1-e^(-"gt"//2v_(0))]`