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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=k2v0hv20+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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