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(C) Let $r$ be the radius of the sphere. Given diameter $D = 42 \, m$,so $r = 21 \, m$.When the particle loses contact,the normal force $N = 0$.At thi

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$35$

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(C) Let $r$ be the radius of the sphere. Given diameter $D = 42 \, m$,so $r = 21 \, m$.When the particle loses contact,the normal force $N = 0$.At this point,the component of gravity provides the centripetal force: $mg \cos 	heta = \frac{mv^2}{r}$,so $v^2 = rg \cos 	heta$.By conservation of energy from the top of the sphere to the point of contact loss:$mg(2r) = mg(r + r \cos 	heta) + \frac{1}{2}mv^2$.Substituting $v^2 = rg \cos 	heta$:$2mgr = mgr(1 + \cos 	heta) + \frac{1}{2}mgr \cos 	heta$.$2 = 1 + \cos 	heta + \frac{1}{2} \cos 	heta$.$1 = \frac{3}{2} \cos 	heta \implies \cos 	heta = \frac{2}{3}$.The height from the center is $x = r \cos 	heta = r(\frac{2}{3}) = \frac{2}{3}r$.The height from the bottom is $h = r + x = r + \frac{2}{3}r = \frac{5}{3}r$.$h = \frac{5}{3} 	imes 21 = 35 \, m$.

$A$ particle starts sliding from the top of a sphere of diameter $42 \, m$. At what height from the bottom of the sphere will the particle lose contact with the sphere?

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