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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. |
Answer» 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))]` |
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