Question

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

• A. $10.6m$
• B. $12.8m$
• C. $15.4m$
• D. $24.2m$

Answer 1

The correct option is C $15.4m$

$\therefore v=v_x\hat{i}+v_y\hat{j}$

$v=i\times u_x\cos \theta \hat{i}+u_x\times \cos \theta \times \tan \left(\theta /2\right)\hat{j}$

$v^2=u^2{\cos }^2\theta +u^2{\cos }^2\theta \times {\tan }^2\theta /2$

$v^2=u^2{\cos }^2\theta (1+{\tan }^2\theta /2)$

$v^2=u^2{\cos }^2\theta \times {\sec }^2\theta /2$

$\therefore$ Radius of curvature $=\frac{u^2{\cos }^2\theta \times {\sec }^2\theta /2}{g\cos \theta /2}$

$=\frac{400\times 1\times 4\times 2}{4\times 3\times 10\times \sqrt{3}}$

$\frac{80}{3\sqrt{3}}$