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Motion In A Plane

Question
CBSEENPH11026344

What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle θ2 with the horizontal?

  • r = v2 cos2θg cos2 θ2

  • r = 2 v sinθg tanθ

  • r =v cosθg sin2θ2

  • r  = 3v cos θg cot θ

Solution

A.

r = v2 cos2θg cos2 θ2

Let v be the velocity at the point where it makes an θ2angle with the horizontal.

The horizontal component remains unchanged. So

   

          vm cosθ2 = u cosθ

⇒                  v = u cosθcosθ2                      .....(i)

From figure 

             mg cosθ2 = m v2r

                 r = v2g cos θ2 

Putting the value of v from Eq (i), we get

             r = v2 cos2θ g cos2 θ2