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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-13/2

c

tan-14/9

d

sin-14/9

answer is C.

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Detailed Solution

H=u2sin2θ2g or 80=u2sin2θ2×10or u2sin2θ=1600 or u sinθ =40 ms-1Horizontal velocity = u cosθ=3 × 30=90 ms-1usinθucosθ=4090or tanθ=49or θ=tan−149
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