Question

A particle is projected from ground with initial speed $u$ and projection angle $\theta$ from horizontal. Find the radius of curvature of trajectory of particle when during flight particle's velocity makes an angle $\frac{\theta }{2}$ with horizontal.

• A. $\frac{u^2}{g}co{s}^2\theta sec\frac{\theta }{2}$
• B. $\frac{u^2}{2g}co{s}^2\theta se{c}^3\frac{\theta }{2}$
• C. $\frac{u^2}{g}co{s}^2\theta se{c}^2\frac{\theta }{2}$
• D. $\frac{u^2}{g}co{s}^2\theta se{c}^3\frac{\theta }{2}$

Answer 1

The correct option is D $\frac{u^2}{g}co{s}^2\theta se{c}^3\frac{\theta }{2}$

During flight, $ucos\theta =vcos\frac{\theta }{2}$

$v=\frac{ucos\theta }{cos\frac{\theta }{2}}$

Normal acceleration of particle at point P is $a_n=gcos\frac{\theta }{2}$

$a_n=\frac{v^2}{r}$

Here, r = radius of curvature

$r=\frac{v^2}{a_n}$

$r=\frac{u^2}{g}co{s}^2\theta se{c}^3\frac{\theta }{2}$