A balloon starts rising from the surface of the earth with a verticle component of velocity `v_(0)`. The baloon gathes a horizontal velocity `v_(x)=lambda_(y)`, where `lambda` is a constant and `y` is the height from the surface of earth, due to a horizontaln wind. determine (a) the equation of trajectory and (b) the tangential, normal and total acceleration of the balloon as a function of `y`.

Explanation

(a) `v_(y0=v_(0)implies(dy)/(dt)=v_(0)`
`int_(0)^(y)dy=v_(0)int_(0)^(t)dtimpliesy=v_(0)t`
`v_(x)=lambdayimplies(dx)/(dt)=lambdav_(0)t` ….`(i)`
`int_(0)^(x)dx=lambdav_(0)int_(0)^(t)t dtimpliesx=(lambdav_(0)t^(2))/(2)` ……`(ii)`
Eliminating `t` between `(i)` and `(ii)`, we get
`x=(lambdayv_(0))/(2)((y)/(v_(0)))^(2)` `implies x=(lambday^(2))/(2v_(0))` (parabola, open rightward)
(b) `v_(y)=v_(0)impliesa_(y)=(dv_(y))/(dt)=0`
`v_(x)=lambday=lambdav_(0)timpliesa_(x)=(dv_(x))/(dt)=lambdav_(0)`
Total acceleration `a=a_(x)=lambdav-(0)`

`x=(lambdaY^(2))/(2V_(0))impliesdx=(lambda)/(2v_(0))2ydy`
Slope: `(dy)/(dx)=tan theta=(v_(0))/(lambday)`
Normal acceleration
`a_(n)=a sin theta=(lambdav_(0)xxv_(0))/sqrt(v_(0)^(2)+lambda^(2)y^(2))=(lambdav_(0)^(2))/sqrt(v_(0)^(2)+lambda^(2)y^(2))`