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Q.

A ball thrown down the incline strikes at a point on the incline 25 m below the horizontal as shown in the figure. If the ball rises to a maximum height of 20 m above the point of projection, the angle of projection α (with horizontal X-axis) is

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a

tan−1⁡43

b

tan−1⁡34

c

tan−1⁡32

d

tan−1⁡23

answer is A.

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Detailed Solution

y=xtan⁡θ−gx22u2cos2⁡θ     −25=75tan⁡α−5(75)2u2cos2⁡α  …….(i)u2sin2⁡α2g=20    ……..(ii)From Eqs. (i) and (ii), we get⇒−25=75tan⁡α−5(75)2sin2⁡α400cos2⁡α⇒4516tan2⁡α−3tan⁡α−1=0⇒45tan2⁡α−48tan⁡α−16=0⇒tan⁡α=43
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