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A solid spherical ball is rolled up an inclined plane of angle of inclination $30^{\circ}$ with an initial speed of $4 \mathrm{~m} / \mathrm{s}$ at the bottom of the inclination. How far will the ball go up the plane. $\left(\right.$ Use $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$
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224 cm
$\begin{aligned} & \text { As P.E } \mathrm{E}_{\mathrm{i}}+(\mathrm{K} . \mathrm{E})_{\mathrm{i}}=(\mathrm{P} . \mathrm{E})_{\mathrm{f}}+(\mathrm{K} . \mathrm{E})_{\mathrm{f}} \\ & \Rightarrow \quad 0+\frac{1}{2} \times \mathrm{m} \times 4^2+\frac{1}{2}\left(\frac{2}{5} \mathrm{mR}^2\right) \times \frac{4^2}{\mathrm{R}^2}=\mathrm{mgh} \\ & \Rightarrow \quad \frac{4^2}{2}+\frac{1}{5} \times 4^2=10 \mathrm{~h} \Rightarrow 8+3.2=10 \mathrm{~h} \\ & \Rightarrow \quad 1.12=\mathrm{h} \Rightarrow \mathrm{h}=1.12 \mathrm{~m} \\ & \text { So, } \ell=\frac{\mathrm{h}}{\sin 30^{\circ}}=\frac{\mathrm{h}}{1 / 2}=2 \mathrm{~h}=2.24 \mathrm{~m}=224 \mathrm{~cm}\end{aligned}$
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