Question

A particle is projected with a speed u at an angle θ with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circular circle? This radius is called the radius of curvature of the curve at the point.

Answer 1

At the highest point, the vertical component of velocity is zero.
So, at the highest point, we have:
velocity = v = ucosθ
Centripetal force on the particle = $\frac{m{v}^2}{r}$
$\Rightarrow \frac{m{v}^2}{r}=\frac{m{u}^2{\cos }^2\mathrm{\theta }}{r}$
At the highest point, we have:
$mg=\frac{m{v}^2}{r}$
Here, r is the radius of curvature of the curve at the point.
$\Rightarrow r=\frac{u^2{\cos }^2\theta }{g}$