Question

Radius of curvature of projectile at point $A$ is?

Answer 1

Step 1: Given information

From the figure,

1. We have to find out the radius of curvature of a projectile at point $A$.
2. It is already given that the projectile is fired with a velocity $v$, making an angle $\alpha$ with the horizontal.
3. At this point, the projectile has an acceleration $g$, which is the acceleration due to gravity.

Step 2: Formula for calculating the radius of curvature:

1. The radius of curvature of a path at a point in a circle tells us how much the curve is at that point.
2. The less the radius of curvature, the more pointed the curve .
3. If the radius of curvature is infinite then it means that the curve is a straight line.
4. Here we have to calculate the radius of curvature (R) for a projectile at its highest point from the ground.
5. At the highest point, the vertical component of velocity of the projectile is $0$ and it has velocity only along with the horizontal direction i.e. $v\cos \alpha$.

The relation between acceleration,$a$(centripetal acceleration), radius,$R$ and velocity $v$ of the body in a uniform circular motion can be given by:

$$\begin{gathered} a=\frac{v^2}{R} \\ \Rightarrow R=\frac{v^2}{a} \\ \Rightarrow R=\frac{v^2}{g} \end{gathered}$$

Where, $g$ is acceleration due to gravity

Step 3: Calculating the radius of curvature
By Substitute the given value of velocity, $v=v\cos \alpha$ in the given equation, we get

The radius of curvature at point $A$,
$$\begin{gathered} R_A=\frac{{(v\mathrm{cos\alpha })}^2}{g} \\ \Rightarrow R_A=\frac{{v_0}^2{\cos }^2\alpha }{g} \end{gathered}$$

Hence this is the required radius of curvature of a projectile at point $A$.