Question

A particle is projected with a speed $u$ at an angle $\theta$ 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 circle? This radius is called the radius of curvature of the curve at the point.

• A. $\frac{u^2{\cos }^2\theta }{2g}$
• B. $\frac{u^2{\cos }^2\theta }{g}$
• C. $\frac{2u^2{\cos }^2\theta }{g}$
• D. $\frac{4u^2{\cos }^2\theta }{g}$

Answer 1

The correct option is B $\frac{u^2{\cos }^2\theta }{g}$

In a projectile (as of given particle) motion , at highest position , we have only a constant horizontal velocity, given by:

$u_x=u\cos \theta$

At highest point, necessary centripetal acceleration is provided by gravitational acceleration $g$, therefore

centripetal acceleration $=g$

or $\frac{(u\cos \theta {)}^2}{r}=g$

or $r=\frac{u^2\cos {\theta }^2}{g}$