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Feynman Fan posted an Question
August 10, 2020 • 19:42 pm 30 points
  • IIT JAM
  • Physics (PH)

A particle is projected with a speed u at an angle theta with horizontal what is the radius of curvature of the parabola traced out by the particle at a point

A particle is projected with a speed u at an angle theta with horizontal what is the radius of curvature of the parabola traced out by the particle at a point where particle makes angle theta/2 with horizontal

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    Ruby negi

    Now, We know that in case of the Projectile motion,  Horizontal velocity remains the constant throughout the motion. ∴ Horizontal velocity at first = u cosθ At Highest Point, Resultant velocity = u cosθ [Since, vertical will be zero.] Now, If the velocity vector is making an angle θ/2 with the horizontal at some point in its trajectory, then Horizontal component = vcosθ/2 ∴  vcosθ/2 = ucosθ ⇒ v = ucosθ/cosθ/2 ∴ mv²/r = mg ⇒ r = v²/g ⇒ r = u²cos²θ/gcos²θ/2 .... hope this will help you..

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