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