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Question

# 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 a horizontal velocity component Vx=Ky where K is a constant and y is the height of ascent. The tangential and normal acceleration of the balloon at a height of ascent equal to h is

A
at=k2v0hv20+k2h2;an=kv2v20+k2h2
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B
at=2k2v0hv20+k2h2;an=2kv2v20+k2h2
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C
at=k2v0hv20+k2h2;an=2kv2v20+k2h2
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D
at=2k2v20v20+k2h2;an=2kv0v20+k2h2
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Solution

## The correct option is A at=k2v0h√v20+k2h2;an=kv2v20+k2h2→anet=ax^i+ay^j ax=dvxdt=kdydt (from equation (ii)) ay=dvydt=0 ⇒→anet=kv0^i [using equation (i) and (ii)] Let θ be the angle between velocity vector and x - axis. This means tanθ is the slop of the trajectory of the balloon. Hence, tanθ=dydx Since, x=ky22v0 ⇒dydx=v0ky at=anetcosθ an=anetsinθ Put anet, cosθ and sinθ to get the result.

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