A balloon starts rising from the earth's surface. The ascension rate is constant and equal to `v_(0)`. Due to the wind. The balloon gathers the horizontal velocity component `v_(x)=ky`, where k is a constnat and y is the height of ascent. Find how the following quantities depednd on the height of ascent.
(a) the horizontal drift of the balloon x (y)
(b) the total tangential and normal accelrations of the balloon.

Explanation

(a) `(dy)/(dt)=v_(0)` .............(i)
`(dx)/(dt)=ky` ............(ii)
Dividing Eqn. (i) by Eqn. (ii) we get
`(dy)/(dx)=(v_(0))/(ky)`
`dx=(ky)/(v_(0))dy`
Integrating we get `x=(k)/(v_(0))(y^(2))/(2)`
This is the desired trajector of the particle, which is and equation of a parabola.
(b) For finding the tangential and normal accelrations, we require and expression for the speed as a function of height y
`v_(y)=v_(0)` and `v_(x)=ky`
`v=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(v_(0)^(2)+k^(2)y^(2))`
Therefore, tangential accelration,
`a_(t)=(du)/(dt)=(k^(2)y)/(sqrt(v_(0)^(2)+k^(2)y^(2))(dy)/(dt)`
`=(k^(2)yv_(0))/(sqrt(v_(0)^(2)+k^(2)y^(2)))`
`a_(t)=(k^(2)y)/(sqrt(1+k^(2)y^(2)//v_(0)^(2)))`
Now, the total accelration is,
`a=sqrt(((dv_(y))/(dt)))^(2)+((dv_(x))/(dt))^(2)`
`=(dv_(x))/(dt)=k(dy)/(dt)=kv_(0)`
Normal accleration, `a_(n)=sqrt(a^(2)-a_(t)^(2))=(kv_(0))/(sqrt(1+((ky)/(v_(0)))^(2))`