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(A) At the highest point,the final kinetic energy $KE_f = 0$.Initial kinetic energy $KE_i$ is the sum of translational and rotational kinetic energy:$

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$75$

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(A) At the highest point,the final kinetic energy $KE_f = 0$.Initial kinetic energy $KE_i$ is the sum of translational and rotational kinetic energy:$KE_i = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$For a hollow spherical ball,the moment of inertia $I = \frac{2}{3} mR^2$. In the case of pure rolling,$v = R\omega$,so $\omega = \frac{v}{R}$.Substituting these values:$KE_i = \frac{1}{2} mv^2 + \frac{1}{2} 	imes (\frac{2}{3} mR^2) 	imes (\frac{v}{R})^2$$KE_i = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 = \frac{5}{6} mv^2$Applying the law of conservation of energy,the initial kinetic energy is converted into potential energy at the maximum height $h$:$KE_i = PE_f$$\frac{5}{6} mv^2 = mgh$$h = \frac{5v^2}{6g}$Given $v = 3\, m/s$ and taking $g = 10\, m/s^2$:$h = \frac{5 	imes (3)^2}{6 	imes 10} = \frac{5 	imes 9}{60} = \frac{45}{60} = 0.75\, m$Converting to centimeters: $0.75\, m = 75\, cm$.

$A$ hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3\, m/s$ (as shown in figure). The maximum height with respect to the initial position covered by it will be $...........cm$.

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