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Question

An object is thrown horizontally with a velocity of 10 m/s. Find the radius of curvature of its trajectory 3 s after the motion has begun. Take g=10 m/s2

A
10010 m
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B
316 m
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C
1005 m
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D
300 m
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Solution

The correct option is A 10010 m
For the given horizontal projection:
ux=10 m/s, uy=0, ax=0, ay=+g
(taking downwards as +ve)

Horizontal velocity does not change. Hence velocities after t=3 s are,
vx3=10 m/svy3=uy+at=0+10×3=30 m/s

Magnitude of velocity after t=3 s is:
v3=v2x3+v2y3=102+302=100+900=1000=1010 m/s


Magnitude of radial acceleration of ball at the point where v3 serves as tangential velocity is
ar=gcosθ

In triangle ABC,
cos θ=ABAC=vx3v3=101010=110
Therefore, ar=10×110=10 m/s2
and Radius of curvature
r=v23ar=100010=10010 m

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