Let at any time t, the velocity vector be inclined at an angle θ with horizontal at a point, say, P as shown in Figure .

Since gravitational acceleration is always acting vertically downwards, the components of perpendicular to the velocity vector can be treated as a radial acceleration towards, the centre of a circular path of radius (let r) as shown in Figure; r is known as radius of curvature of the parabola at P.

$ \displaystyle a_r = \frac{V^2}{r} $

$ \displaystyle r = \frac{V^2}{a_r} $

$ \displaystyle r = \frac{V^2}{g cos\theta} $

where V and θ can be found in terms of t as

$ \displaystyle V = \sqrt{V_0^2 + (gt)^2 – 2 V_0 gt sin \theta_0 } $

Solved Example: A boy is standing inside a train moving with a constant velocity of magnitude 10 m/sec. He throws a ball vertically up with a speed 5 m/sec relative to the train. Find the radius of curvature of the path of the ball just at the time of projection.

Solution: The ball continues to move horizontally with (V_{x})_{0} = 10 m/sec. It begins to move up with (V_{y})_{0} = 5 m/sec. Therefore θ_{0} is given as,

Putting , $ \displaystyle V = \sqrt{10^2 + 5^2} = 5 \sqrt{5} $ m/s

g = 10 m/s^{2}

we obtain , r ≈ 14 m

Solved Example : Referring to the previous illustration, if the boy releases the ball from rest, what will be the radius of curvature of the path at the instant of its release?

Sol: Radius of curvature ,

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

θ = angle between v & horizontal = 0 when the ball is released, it starts moving horizontally with a speed v = speed of the train = 10 m/sec.