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=2v2sinθcosθa
b
h=2v2sinθcosθg
c
h=2v2gsinθcosθ+agsinθ
d
h=2v2asinθcosθ+gasinθ
answer is D.
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Detailed Solution
In X-direction, 0=(vsinθ)t+12(−a)t2⇒t=2vsinθa In Y- direction, h=(vcosθ)t+12gt2⇒h=2v2asinθcosθ+gasinθ