A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to vo. 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 the total, tangential, and normal accelerations of the balloon.
ar=av0;at=a2y√1+a2y2v20;aN=av0 ⎷(1+a2y2v20)
The path of the balloon will look something like this
After t sec ballon would have gone a height of v0tthen at that very instance the balloon's vx will be
vx=av0t⇒ax=av0
Vy=v0
ay=0
ar=total acceleration=av0.......(1)
vtangentical is actually the resultant velocity
⇒vt=√v20+(av0t)2
⇒vt=v0√1+a2t2
at=dvtdt=v0a2t√1+a2t2⇒a2y√1+(ayv0)2
Now we know a2t+a2N=a2r
⇒v20a4t2(1+a2t2)+a2N=a2v20
aN=√v20a4t2−v20a4t2(1+a2t2)
aN=av0√1+a2t2−a2t2(1+a2t2)
aN=av0√1+a2t2
⇒av0√1+(ayv0)2