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What is the radius of curvature of the p...

What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes and angle `theta/2` with the horizontal?

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The correct Answer is:
A, B, C
Let `v` be the velocity at that instant. Then, horizontal component of velocity remains unchanged.

`:. vcos.(theta)/(2)=ucostheta` or `v=(ucostheta)/(cos.(theta)/(2))`
Tangential component of acceleration of this instant will be,
`a_(t)=gcos(pi//2+theta//2)=-gsintheta//2`
`a_(n)=sqrt(a^(2)-a_(t)^(2))=sqrt(g^(2)-g^(2)sin^(2)(theta)/(2))`
`=gcos(theta)/(2)`
Since, `a_(n)=(v^(2))/(R)`
or `R=(v^(2))/(a_(n))=((ucostheta)/(cos(theta)/(2)))^(2)/(gcos((theta)/(2)))=(u^(2)cos^(2)theta)/(gcos^(3)((theta)/(2)))`
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