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Question

A ball is thrown horizontally from the top of a tower with a speed of 5 m/s . What will be the radius of curvature of its trajectory 1 s after the ball is thrown? Take g=10 m/s2.

A
35210 m
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B
2525 m
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C
25410 m
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D
2545 m
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Solution

The correct option is B 2525 m

For the given horizontal projection:
ux=5 m/s, uy=0, ax=0, ay=+g
Reference Y-axis has been taken vertically downwards as +ve.

Horizontal velocity does not change. Hence velocities after t=1 s are
vx1=5 m/svy1=uy+at=0+10×1=10 m/s

Magnitude of velocity after t=1 s is:
v1=v2x1+v2y1=52+102=25+100=125=55 m/s


If you consider a circle as shown in figure with v1 tangential, magnitude of radial acceleration of ball,
ar=gcosθ

From the shown figure of vector triangle ;
cos θ=vx1v1=555
Therefore,
ar=10×555=25 m/s2
Radius of curvature
r=v21ar=12525=2525 m

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