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Question

# 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.

A

ar=av0;at=av01+(ayv0)2;aN=av0

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B

ar=av0;at=av0;aN=0

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C

ar=av0;at=a2y1+a2y2v20;aN=av0 (1+a2y2v20)

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D

ar=a2y1+a2y2v20;at=av0;aN=av0

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Solution

## The correct option is C 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

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