An object is thrown horizontally with a velocity of 10m/s. Find the radius of curvature of its trajectory 3s after the motion has begun. Take g=10 m/s2
A
100√10m
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B
316m
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C
100√5m
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D
300m
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Solution
The correct option is A100√10m For the given horizontal projection: ux=10m/s,uy=0,ax=0,ay=+g
(taking downwards as +ve)
Horizontal velocity does not change. Hence velocities after t=3s are, vx3=10m/svy3=uy+at=0+10×3=30m/s
Magnitude of velocity after t=3s is: v3=√v2x3+v2y3=√102+302=√100+900=√1000=10√10m/s
Magnitude of radial acceleration of ball at the point where →v3 serves as tangential velocity is ar=gcosθ
In triangle ABC, cosθ=ABAC=vx3v3=1010√10=1√10
Therefore, ar=10×1√10=√10m/s2
and Radius of curvature r=v23ar=1000√10=100√10m