Question

A ball is projected with initial speed u and making an angle $\theta$ with the vertical. Consider a small part of the trajectory near the highest position and take it approximately to be a circular arc. What is the radius of this circle? [This radius is called the radius of curvature of the curve at the point].

Answer 1

Given,

Initial velocity $=u$

The angle of projection $=\theta$

Components of initial velocity and acceleration

In vertical direction, $v_y=u\sin \theta ,a_y=-g$

In horizontal direction, $v_x=u\cos \theta ,a_x=0$

Point P, the height of the projectile curve is maximum.

At point P, Components of velocity and acceleration

In vertical direction, $v_y=0,a_y=-g$

In horizontal direction, $v_x=u\cos \theta ,a_x=0$

Direction of centripetal acceleration $a_c$, is normal to tangential velocity $v$ .

Centripetal acceleration at point P, is gravity, $a_c=g$ .....(1)

Centripetal acceleration in form of tangential velocity, $a_c=\frac{v^2}{r}=\frac{{\left(u\cos \theta \right)}^2}{r}$ ......(2)

Equating both equation (1) and (2)

$\Rightarrow g=\frac{{\left(u\cos \theta \right)}^2}{r}$

$\Rightarrow r=\frac{{\left(u\cos \theta \right)}^2}{g}$

$\therefore$ Radius of curvature at highest $\Rightarrow r=\frac{{\left(u\cos \theta \right)}^2}{g}$