Question

A particle is projected with speed $u$ at an angle $\theta$ with the horizontal. Find the radius of curvature at the highest point of its trajectory.

• A. $\frac{u^2{\cos }^2\theta }{2g}$
• B. $\frac{\sqrt{3}{u}^2{\cos }^2\theta }{2g}$
  
• C. $\frac{u^2{\cos }^2\theta }{g}$
• D. $\frac{\sqrt{3}{u}^2{\cos }^2\theta }{g}$

Answer 1

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


We know that, the horizontal component of velocity remains constant and at highest point, the vertical component of velocity is zero. Acceleration will be equal to g and perpendicular to velocity.
At highest point, $mg=\frac{m{v}^2}{r}$, where $r$ is the radius of curvature
$\Rightarrow r=\frac{v^2}{g}=\frac{(u\cos \theta {)}^2}{g}$
$\Rightarrow r=\frac{u^2{\cos }^2\theta }{g}$
Option c is correct.