A cylinder is placed on a horizontal surface. A particle is projected with speed u and it crosses the cylinder by just touching at two points as shown in figure. Then the initial speed with which the particle projected is $\sqrt{6 x} m / s$ , Find x. $(g = 10 m / s^{2})$

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answer is 5.

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Detailed Solution

$2 R \sin 60^{0} = \frac{v_{0}^{2} \sin 120^{0}}{g}$

$\begin{array}{l} v_{0}^{2} = 2 R g = 2 (0.6) \times 10 = 12 \\ v_{0}^{2} = u^{2} - 2 g (\frac{3 R}{2}) \\ u^{2} = 12 + 10 \times 3 \times 0.6 = 30 \\ u = \sqrt{30} \end{array}$

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