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Q.

A particle is ejected from the tube at A with a velocity v at an angle θ with the vertical y-axis. A strong horizontal wind gives the particle a constant horizontal acceleration a in the x-direction. If the particle strikes the ground at a point directly under its released position and the downward y-acceleration is taken as g then

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a

h=2v2sin⁡θcos⁡θa

b

h=2v2sin⁡θcos⁡θg

c

h=2v2gsin⁡θcos⁡θ+agsin⁡θ

d

h=2v2asin⁡θcos⁡θ+gasin⁡θ

answer is D.

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Detailed Solution

In X-direction, 0=(vsin⁡θ)t+12(−a)t2⇒t=2vsin⁡θa In Y- direction,  h=(vcos⁡θ)t+12gt2⇒h=2v2asin⁡θcos⁡θ+gasin⁡θ
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