Question

A particle is projected from the ground at an angle 60° with the horizontal that speed u= 20m/s find the radius of curvature of the path of the particle when its velocity makes an angle of 30° with horizontal

Answer 1

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Answer 2

Answer:

The radius of curvature of the path of the particle is 15.39 m.

Explanation:

Given that.

Angle = 60°

Speed u = 20 m/s

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

$v cos 30^{\circ}=u\ cos\ 60^{\circ}$

$v = \dfrac{u\ cos\ 60^{\circ}}{cos\ 30^{\irc}}$

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

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

The radius of curvature is

$g\ cos\ 30^{\circ}=\dfrac{v^2}{R}$

$R = \dfrac{v^2}{g\ cos\ 30^{\circ}}$

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

$R = \dfrac{80}{3\sqrt{3}}$

$R = 15.39\ m$

Hence, The radius of curvature of the path of the particle is 15.39 m.