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(B) Applying the law of conservation of energy between the top of the track $(A)$ and the horizontal section $(BC)$:$mgh_A = mgh_B + \frac{1}{2}mv^2 +

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$2$

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(B) Applying the law of conservation of energy between the top of the track $(A)$ and the horizontal section $(BC)$:$mgh_A = mgh_B + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since the sphere rolls without slipping,$I = \frac{2}{5}mR_s^2$ and $\omega = \frac{v}{R_s}$,where $R_s$ is the radius of the sphere.$mg(h_A - h_B) = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR_s^2)(\frac{v^2}{R_s^2})$$mg(2.4 - 1.0) = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$$mg(1.4) = \frac{7}{10}mv^2$$v^2 = \frac{1.4 	imes 10 	imes g}{7} = 2g = 20 \ m^2/s^2 \implies v = \sqrt{20} \ m/s$.Now,for the projectile motion from point $C$ to the ground:The height $h = 1.0 \ m$. The time taken to fall is $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 	imes 1.0}{10}} = \sqrt{0.2} \ s$.The horizontal distance $R = v 	imes t = \sqrt{20} 	imes \sqrt{0.2} = \sqrt{4} = 2 \ m$.

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