Question

A particle is projected with a speed $u$ at an angle $\theta$ with the horizontal. Find the radius of curvature of the parabola traced out by the projectile at a point, where the particle velocity makes an angle $\theta /2$ with the horizontal.

• A. $\frac{u^2{\cos }^2\theta }{g{\cos }^3\theta }$
• B. $\frac{{\cos }^2\theta }{g{\cos }^3\left(\theta /2\right)}$
• C. $\frac{u^2{\cos }^2\theta }{g{\cos }^3\left(\theta /2\right)}$
• D. $\frac{u^2{\cos }^2\theta }{{\cos }^3\left(\theta /2\right)}$

Answer 1

The correct option is C $\frac{u^2{\cos }^2\theta }{g{\cos }^3\left(\theta /2\right)}$

From figure, $u_x=ucos\theta$

x direction : $a_x=0$ $⟹V_x=u_x=ucos\theta$

Also $tan\left(\theta /2\right)=\frac{V_y}{V_x}$

$\therefore$ $V_y=V_{x}tan\left(\theta /2\right)=ucos\theta$ $tan\left(\theta /2\right)$

Speed at that instant $V=\sqrt{V_x^2+V_y^2}=\sqrt{(ucos\theta {)}^2+[ucos\theta tan\left(\theta /2\right){]}^2}=\frac{ucos\theta }{cos\left(\theta /2\right)}$

Also from figure, $a_r=acos\left(\theta /2\right)$

$\therefore$ $a_r=gcos\left(\theta /2\right)$

Let the radius of curvature at that instant be $R$

$\therefore$ $R=\frac{V^2}{a_r}=\frac{\frac{u^{2}co{s}^2\theta }{co{s}^2\left(\theta /2\right)}}{gcos\left(\theta /2\right)}=\frac{u^{2}co{s}^2\theta }{gco{s}^3\left(\theta /2\right)}$