Question

A particle is projected from the ground at an angle of ${60}^{\circ }$ with the horizontal with speed $u=20\text{m/s}$. The radius of curvature of the path of the particle, when its velocity makes an angle ${30}^{\circ }$ with the horizontal is
$(g=10{\text{m/s}}^2)$

• A. $10.6\text{m}$
• B. $12.8\text{m}$
• C. $15.4\text{m}$
• D. $24.2\text{m}$

Answer 1

The correct option is C $15.4\text{m}$

Let $v$ be the velocity of particle when it makes ${30}^{\circ }$ with the horizontal.


Since the horizontal component doesn't change during projectile motion,
$v\cos {30}^{\circ }=u\cos {60}^{\circ }$
$\Rightarrow v=\frac{u\cos {60}^{\circ }}{\cos {30}^{\circ }}$
$=\frac{\left(20\right)\left(1/2\right)}{\left(\sqrt{3}/2\right)}=\frac{20}{\sqrt{3}}\text{m/s}$

Now, centripetal acceleration at the point $=g\cos {30}^{\circ }=\frac{v^2}{R}$
where $R$ is the radius of curvature.

$\Rightarrow R=\frac{v^2}{g\cos {30}^{\circ }}=\frac{(20/\sqrt{3}{)}^2}{10\left(\sqrt{3}/2\right)}=15.4\text{m}$

Hence, the correct answer is option (c).