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Question

A particle is projected from the ground at an angle of 60o with horizontal with a speed u=20m/s. The radius of curvature of the path of the particle, when its velocity makes an angle of 30o with the horizontal is (g=10m/s2)

A
10.6 m
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B
12.8 m
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C
15.4 m
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D
24.2 m
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Solution

The correct option is C 15.4 m
v=vx^i+vy^j
v=i×uxcosθ^i+ux×cosθ×tan(θ/2)^j
v2=u2cos2θ+u2cos2θ×tan2θ/2
v2=u2cos2θ(1+tan2θ/2)
v2=u2cos2θ×sec2θ/2
Radius of curvature =u2cos2θ×sec2θ/2gcosθ/2
=400×1×4×24×3×10×3
8033

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