Question

What is the radius of curvature of the parabola traced out by the projectile projected at a speed $v$ and projected at an angle $\theta$ with the horizontal at a point where the particle velocity makes an angle $\frac{\theta }{2}$ with the horizontal?

• A. $r=\frac{v^{2}co{s}^2\theta }{gco{s}^3\frac{\theta }{2}}$
• B. $r=\frac{2vsin\theta }{gtan\theta }$
• C. $r=\frac{vcos\theta }{gsi{n}^2\frac{\theta }{2}}$
• D. $r=\frac{3vcos\theta }{gcot\theta }$

Answer 1

The correct option is A $r=\frac{v^{2}co{s}^2\theta }{gco{s}^3\frac{\theta }{2}}$

Body is projected with a speed $v$ at an angle $\theta$ with the horizontal.

At the point of interest, angle with horizontal is $\theta /2$

$\tan \left(\theta /2\right)=v_y/v_x............\left(i\right)$

$v^2=v_x^2+v_y^2$

Centripetal acceleration is:

$a_c=\frac{v^2}{R}=g\cos \left(\theta /2\right)$

$R=\frac{v_x^2+v_y^2}{g\cos \left(\theta /2\right)}$

Using (i), $R=\frac{v_x^2(1+{\tan }^2(\theta /2\left)\right)}{g\cos \theta /2}$

We know, $v_x=v\cos \theta$

and $se{c}^2\theta =1+{\tan }^2\theta$

$\therefore R=\frac{v^2{\cos }^2\theta {\sec }^2\left(\theta /2\right)}{g\cos \theta /2}$

$R=\frac{v^2{\cos }^2\theta }{g{\cos }^3\left(\theta /2\right)}$

Note- This can be also solved by dimensional analysis