What is the radius of curvature of the parabola traced out by the...

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 an angle

\frac{\theta}{2}

with the horizontal? What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where

Options

A $r=\frac{v^2 \cos ^2 \theta}{g \cos ^2 \frac{\theta}{2}}$
B $r=\frac{2 v \sin \theta}{g \tan \theta}$
C $r=\frac{v \cos \theta}{g \sin ^2 \frac{\theta}{2}}$
D $r=\frac{3 v \cos \theta}{g \cot \theta}$

Correct answer

A. $r=\frac{v^2 \cos ^2 \theta}{g \cos ^2 \frac{\theta}{2}}$

Step-by-step solution

Let

v

be the velocity at the point where it makes an angle

\frac{\theta}{2}

with the horizontal. The horizontal component remains unchanged.So, What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where

v_m \cos \left(\frac{\theta}{2}\right)=u \cos \theta

\Rightarrow \quad v=\frac{u \cos \theta}{\cos \left(\frac{\theta}{2}\right)}

\ldots

(i) From figure,

m g \cos \left(\frac{\theta}{2}\right)=\frac{m v^2}{r}

r=\frac{v^2}{g \cos \frac{\theta}{2}}

Putting the value of

v

from Eq. (i), we get

r=\frac{v^2 \cos ^2 \theta}{g \cos ^2 \frac{\theta}{2}}

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