Question

A body is projected with a speed $u$ at an angle $\theta$ with the horizontal. The radius of curvature of the trajectory, when it makes an angle $\left(\frac{\theta }{2}\right)$ with the horizontal is $(g$ -acceleration due to gravity)

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

Answer 1

Answer: (D)

Let velocity of projectile is $v$ at an angle $\frac{\theta }{2}$ with horizontal
$\therefore v\cos \frac{\theta }{2}=u\cos \theta$
or $v=\frac{u\cos \theta }{\cos \frac{\theta }{2}}$

As horizontal component remains same.
Also, centripetal force is provided by the component of weight.
So, $\frac{m{v}^2}{r}=mg\cos \frac{\theta }{2}$
Hence, radius of curvature of path,
$\Rightarrow r=\frac{v^2}{g\cos \frac{\theta }{2}}$
$=\frac{\frac{u^2{\cos }^2\theta }{{\left(\cos \frac{\theta }{2}\right)}^2}}{g\cos \frac{\theta }{2}}$
$\Rightarrow r=\frac{u^2{\cos }^2\theta \cdot {\sec }^3\left(\frac{\theta }{2}\right)}{g}$