Question

If a particle is projected with speed, $u$, from ground at an angle, $\theta$, with horizontal, then radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by:

• A. $\frac{u^{2}co{s}^2\theta }{g}$
• B. $\frac{u^{2}co{t}^2\theta }{gsin\theta }$
• C. $\frac{u^2}{g}$
• D. $\frac{u^{2}ta{n}^2\theta }{gcos\theta }$

Answer 1

Answer: (B)

$\frac{u^{2}co{t}^2\theta }{gsin\theta }$

Let the velocity at the required point be $V$.

$U_x=U\cos \theta U_y=U\sin \theta V_x=V\cos \alpha V_y=V\sin \alpha$

and $V_x=U_x$ Since there is no acceleration in $x$ direction. So we have

$V\cos \alpha =U\cos \theta \Rightarrow V=\frac{U\cos \theta }{\cos \alpha }$

By using trigonometry, we get $\alpha ={90}^o-\theta$

$\Rightarrow V=\frac{U\cos \theta }{\cos \alpha }=\frac{U\cos \theta }{\sin \theta }=U\cot \theta$

Centripetal force $=\frac{m{V}^2}{r}=mg\cos \alpha =mg\sin \theta$ $(\because \alpha ={90}^o-\theta )$

$\Rightarrow r=\frac{V^2}{g\sin \theta }=\frac{U^2{\cot }^2\theta }{g\sin \theta }$