A particle is projected from the ground at an angle of $60{^\circ}$ with horizontal with speed $u = 20\ ms^{- 1}$. The radius of curvature of the path of the particle, when its velocity makes an angle of $30{^\circ}$ with horizontal is $(g\ = 10\ ms^{- 2})$

Question

Dr. Abubacker · Accepted Answer

(c) Let $v$ be the velocity of particle when it makes $30{^\circ}$ with horizontal. Then
[image:4_image148.emf]
$v\cos{30{^\circ} =}u\cos{60{^\circ}}$
or $v = \frac{{u\cos}{60{^\circ}}}{\cos{30{^\circ}}} = \frac{20\left( \frac{1}{2} \right)}{\left( \frac{\sqrt{3}}{2} \right)} = \frac{20}{\sqrt{3}}ms^{- 1}$
Now, $g\cos{30{^\circ}} = \frac{v^{2}}{R}$
or $R = \frac{v^{2}}{g\cos{30{^\circ}}} = \frac{\left( \frac{20}{\sqrt{3}} \right)^{2}}{(10)\ \frac{\sqrt{3}}{2}}$
= 15.4 m

Q. A particle is projected from the ground at an angle of $60{^\circ}$ with horizontal with speed $u = 20\ ms^{- 1}$. The radius of curvature of the path of the particle, when its velocity makes an angle of $30{^\circ}$ with horizontal is $(g\ = 10\ ms^{- 2})$

Explanation

Reviewed by Dr. Abubacker

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