A tennis ball (treated as a hollow spherical | vedclass

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(B) The tennis ball rolls down from height $H = 2.0 \ m$ to a point $A$ at height $h = 0.2 \ m$. The vertical drop is $h' = H - h = 2.0 - 0.2 = 1.8 \

Vedclass · Accepted Answer

$2.08$

Vedclass · Answer

(B) The tennis ball rolls down from height $H = 2.0 \ m$ to a point $A$ at height $h = 0.2 \ m$. The vertical drop is $h' = H - h = 2.0 - 0.2 = 1.8 \ m$.Using the law of conservation of energy for a rolling body:$mgh' = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since $I = \frac{2}{3}mR^2$ and $\omega = v/R$:$mgh' = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2$$v^2 = \frac{6}{5}gh' = \frac{6}{5} 	imes 9.8 	imes 1.8 = 21.168 \ m^2/s^2$.The horizontal range $AB$ for a projectile launched at angle $	heta = 30^\circ$ from height $h = 0.2 \ m$ is given by:$AB = \frac{v \cos 	heta}{g} \left( v \sin 	heta + \sqrt{(v \sin 	heta)^2 + 2gh} ight)$$v \sin 30^\circ = \sqrt{21.168} 	imes 0.5 \approx 4.601 	imes 0.5 = 2.3005 \ m/s$$v \cos 30^\circ = 4.601 	imes 0.866 = 3.984 \ m/s$$AB = \frac{3.984}{9.8} \left( 2.3005 + \sqrt{(2.3005)^2 + 2 	imes 9.8 	imes 0.2} ight)$$AB = 0.4065 	imes (2.3005 + \sqrt{5.292 + 3.92}) = 0.4065 	imes (2.3005 + 3.035) = 0.4065 	imes 5.3355 \approx 2.168 \ m$.Given the options,the closest value is $2.08 \ m$.

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