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Sir 6th ano 7th question please Track OABCD (as shown in figure) is smooth and fixed in vertical plane. What minimum speed has to be given to a particle lying at point A so that it can reach point C

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mgh=\frac{1}{2}mv^2\\v=\sqrt{2gh}=\sqrt{2\times10\times320}=80m/s

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Safeer PP

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The speed to reach B is obtained by

\frac{1}{2}mu^2=mgh\ u = initial\ velocity

u=80m/s. A slight push is enough for ball to reach C

so minimum speed is 80m/s

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Safeer PP

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