A ball is thrown horizontally from the top of a tower with a speed of 5m/s . What will be the radius of curvature of its trajectory 1s after the ball is thrown? Take g=10 m/s2.
A
352√10m
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B
252√5m
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C
254√10m
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D
254√5m
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Solution
The correct option is B252√5m
For the given horizontal projection: ux=5m/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=1s are vx1=5m/svy1=uy+at=0+10×1=10m/s
Magnitude of velocity after t=1s is: v1=√v2x1+v2y1=√52+102=√25+100=√125=5√5m/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 ; cosθ=vx1v1=55√5
Therefore, ar=10×55√5=2√5m/s2 ∴ Radius of curvature r=v21ar=1252√5=252√5m