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(B) Let the block leave the hemisphere at point $B$ at a height $h$ from the ground. At this point,the normal reaction $N = 0$.The radial component of

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$\frac{19r}{27}$

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(B) Let the block leave the hemisphere at point $B$ at a height $h$ from the ground. At this point,the normal reaction $N = 0$.The radial component of the gravitational force provides the necessary centripetal force:$mg \cos 	heta = \frac{mv^2}{r}$Since $h = r \cos 	heta$,we have $\cos 	heta = h/r$. Substituting this:$mg(h/r) = \frac{mv^2}{r} \implies v^2 = gh \quad \dots(1)$Now,apply the law of conservation of energy between the top point $A$ (height $r$) and point $B$ (height $h$):$E_A = E_B$$mgr + \frac{1}{2} m u^2 = mgh + \frac{1}{2} mv^2$Given $u = u_0/3$ and $u_0 = \sqrt{gr}$,we have $u^2 = \frac{gr}{9}$.Substituting $u^2$ and $v^2 = gh$ into the energy equation:$mgr + \frac{1}{2} m \left(\frac{gr}{9}ight) = mgh + \frac{1}{2} m(gh)$Divide by $mg$:$r + \frac{r}{18} = h + \frac{h}{2}$$\frac{19r}{18} = \frac{3h}{2}$$h = \frac{19r}{18} 	imes \frac{2}{3} = \frac{19r}{27}$

In the given figure,for an initial velocity $u = u_0/3$,find the height from the ground at which the block leaves the hemisphere,where $u_0 = \sqrt{gr}$.

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