A stone of mass 2 kg is projected from ground with speed 20 m-s at angle 30- with horizontal- The centripetal force acting on the object 1 s after projection is

To find v: v_x=u_x=10√(3) m/s v_y=u_y gt=u sinα gt v_y=20 sin 30^∘ 10=0 So, at t=1 s particle is at maximum height and velocity is horizontal. Force acting pe ...

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Standard XII

Physics

Radius of Curvature

A stone of ma...

Question

A stone of mass $2 kg$ is projected from ground with speed $20 m/s$ at angle $30^{\circ}$ with horizontal. The centripetal force acting on the object $1 s$ after projection is

10 N

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20 N

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30 N

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40 N

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Solution

The correct option is B $20 N$
To find $v :$
$v_{x} = u_{x} = 10 \sqrt{3} m/s$
$v_{y} = u_{y} - g t = u sin α - g t$
$v_{y} = 20 sin 30^{\circ} - 10 = 0$
So, at $t = 1 s$ particle is at maximum height and velocity is horizontal.

Force acting perpendicular to velocity at this instant is $m g$ . This contributes to centripetal force.

$∴ F_{c} = m g = 20 N$

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A stone of mass 2 kg is projected from ground with speed 20 m/s at angle 30∘ with horizontal. The centripetal force acting on the object 1 s after projection is

The correct option is B 20 N To find v: vx=ux=10√3 m/s vy=uy−gt=usinα−gt vy=20sin30∘−10=0 So, at t=1 s particle is at maximum height and velocity is horizontal. Force acting perpendicular to velocity at this instant is mg. This contributes to centripetal force. ∴Fc=mg=20 N

A stone of mass $2 kg$ is projected from ground with speed $20 m/s$ at angle $30^{\circ}$ with horizontal. The centripetal force acting on the object $1 s$ after projection is