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Q.

A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of 3 ms-2 for 0.5 min. If the maximum height reached by it is 80 m, then the angle of projection is (g = 10 ms-2)

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a

tan-13∣

b

tan-1(3/2)

c

tan-1(4/9)

d

sin-1(4/9)

answer is C.

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Detailed Solution

H=u2sin2θ2 g   or   80=u2sin2θ2×10 u2sin2θ=1600   or   usinθ=40 ms-1  Horizontal velocity =ucosθ=3×30=90 ms-1 usinθucosθ=4090 tanθ=49 θ=tan-149
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