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A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to v0. Due to the wind the balloon gathers the horizontal velocity component vx = ay, where a is a constant and y is the height of ascent. Find how the following quantities depend on the height of ascent: (a) the horizontal drift of the balloon x(y); (b) the total, tangential, and normal accelerations of the balloon.

```
6 years ago

```							(a)   When balloon is at sarting pointVy=v°, Vx=0since vertical velocity is constant thus vertical acceleration is 0Now after time t Vy=v°, Vx =ayWhere y =v°t (vertical distance = Vert.velocity*time)Thus Vx=av°t         dx/dt=av°tdx=av°t dtIntegrating both sidesX = av°t2/2X=ay2/2v° = horizontal drift         (since t=y/v°)(b) Now total accr. Is due to VxVx =aydVx/dt= a(dy/dt)Ax= a Vy =         av°        = total accr. (Since Vy =v°)(c) total velocity= ay i + v° j  (in vector form)Tangential accr. = Accr. Along total velocityav° * unit vector along total velocity (dot product)av° i . [(ay i + v°j)/ {(ay)2+(v°)2}1\2 ] .                                   (Since     accr.  dot   total velocity in vector form divided by magnitude of total velocity)After solving you get= a2v°y/(v°2 + a2y2)½  = Tang. accr. (d)  norm. accr. ={ ( Total accr.)2 -  (Tang. accr.)2}½{(av°)2 - (Tang. accr.)2}½do it yourself and get              av°2/{v°2 + a2y2}½.       Thank you...
```
2 years ago
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