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What is the radius of curvature of the parabola traced out by the projectile.Projected with a speed `u=sqrt30` at angle `theta=60^(@)` with the horizontal at a point where the particle velocity makes an angle `theta//2` with the horizontal ? |
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Answer» Correct Answer - A As velocity along horizontal remains constant `U cos theta=V cos (theta/2)` Radius of curvature `r=v^(2)/a_(T)=(v^(2)cos^(2)theta)/(g cos^(2)(theta//2)` |
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