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 θ/2 with the horizontal?

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Solution

Let u be the initial velocity and v be the velocity at the point where it makes an angle $\frac{θ}{2}$ with the horizontal component.
It is given that the horizontal component remains unchanged.
Therefore, we get:
$v \cos (\frac{θ}{2}) = u cosθ$

$\Rightarrow v = \frac{u \cos θ}{\cos \frac{θ}{2}} . . . (i) m g \cos \frac{θ}{2} = \frac{m v^{2}}{r} . . . (ii) \Rightarrow r = \frac{v^{2}}{g \cos \frac{θ}{2}}$
On substituting the value of v from equation (i), we get:
$r = \frac{u^{2} \cos^{2} θ}{g \cos^{2} \frac{θ}{2}}$

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