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Home / Doubts and Solutions/JEE/Physics

a particle is projected with speed 10m/s by making angle 37 with the horizontal. the radius of the curvature at that instant of projection is
A) 12.5 B) 20 C) 15.4 D) 25m

Asked by Tanishq 13th October 2017, 7:20 AM
Answered by Expert
Answer:
Radius of curvature ROC = fraction numerator open square brackets 1 plus open parentheses begin display style fraction numerator d y over denominator d x end fraction end style close parentheses squared close square brackets to the power of begin display style bevelled 3 over 2 end style end exponent over denominator open vertical bar begin display style fraction numerator d squared y over denominator d x squared end fraction end style close vertical bar end fraction
trajectory equation for projectile is given by     y space equals space x tan theta space minus space 1 half fraction numerator g x squared over denominator u squared cos squared theta end fraction
 
hence first and second derivatives are given by

                                   fraction numerator d y over denominator d x end fraction space equals space tan theta space minus space fraction numerator g x over denominator u squared cos squared theta end fraction fraction numerator d squared y over denominator d x squared end fraction space equals space minus fraction numerator g over denominator u squared cos squared theta end fraction
 
Hence ROC = fraction numerator open parentheses 1 plus tan squared theta close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent cross times u squared cross times cos squared theta over denominator g end fraction space equals space fraction numerator u squared over denominator g cos theta end fraction space equals space fraction numerator 10 space cross times space 10 over denominator 9.8 space cross times cos 37 end fraction space equals space 12.78
So the answer ' A) 12.5 ' can be selected
Answered by Expert 13th October 2017, 11:51 AM
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