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Question

A particle is projected from the ground at an angle of 60 with the horizontal with speed u=20 m/s. The radius of curvature of the path of the particle, when its velocity makes an angle of 30 with the horizontal is (g=10 m/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
Radius of curvature =(Velocity)2Radial Acceleration

From the figure, we get
Radial acceleration =gcosθ
Radius of curvature r=v2gcosθ

Since, there is no acceleration along horizontal direction
vcos30=20cos60
v=203 m/s

r=2023×10×32=15.4 m

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