Question

A particle is projected from the ground at an angle 60 degree with the horizontal with speed 20 m per second the radius of curvature of the path of projectile when its velocity makes an angle 30 degree with horizontal is

Answer 1

Answer:

The radius of curvature of the path of projectile  is 15.71 m.

Explanation:

Given that,

Initial velocity = 20 m/s

Let v be the velocity of particle when it makes an angle 30 degree with horizontal.

Since the horizontal velocity does not changes, that is why the initial horizontal component u cos 60° will remain same and will be equal to horizontal component of the final velocity at 30°

$v\ cos30=u\ cos60$

$v = \dfrac{u\ cos60^{\circ}}{cos30^{\circ}}$

$v = \dfrac{20\times\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}$

$v=\dfrac{20}{\sqrt{3}}$

Using centripetal force

$F =\dfrac{mv^2}{R}$

$mg\cos30^{\circ}=\dfrac{mv^2}{R}$

$R = \dfrac{(\dfrac{20}{\sqrt{3}})^2}{9.8\times\dfrac{\sqrt{3}}{2}}$

$R =15.71\ m$

Hence, The radius of curvature of the path of projectile  is 15.71 m.