# A solid sphere of mass 4 kg rolls up a 60° incline with an initial speed of 20 m/s. The maximum height reached by the sphere is (g = 10 m/s2

$\\\text{In case of pure rolling, } \frac{K_{R}}{K_{T}}=\frac{2}{5} \\\therefore Total kinetic energy =7 / 5 times the translational kinetic energy. At highest point, whole of the kinetic energy will be convered into potential energy \\ \therefore m g h=\frac{7}{5}\left(\frac{1}{2} m v^{2}\right) or h=\frac{7 v^{2}}{10 g}=\frac{7(20)^{2}}{10 g}=\frac{7(20)^{2}}{10 \times 10}=28 m$

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