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(A) The horizontal range $R$ of water issuing from a hole at depth $y$ from the surface in a tank of total height $H$ is given by $R = 2\sqrt{y(H-y)}$

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Depends on $H$

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(A) The horizontal range $R$ of water issuing from a hole at depth $y$ from the surface in a tank of total height $H$ is given by $R = 2\sqrt{y(H-y)}$.For hole $A$,the depth from the surface is $h_1$,so the range is $R_A = 2\sqrt{h_1(H - h_1)}$.For hole $B$,the height from the bottom is $h_2$,so the depth from the surface is $(H - h_2)$. The range is $R_B = 2\sqrt{(H - h_2)h_2}$.Since the ranges are equal,$R_A = R_B$:$2\sqrt{h_1(H - h_1)} = 2\sqrt{(H - h_2)h_2}$Squaring both sides:$h_1(H - h_1) = h_2(H - h_2)$$Hh_1 - h_1^2 = Hh_2 - h_2^2$$H(h_1 - h_2) = h_1^2 - h_2^2$$H(h_1 - h_2) = (h_1 - h_2)(h_1 + h_2)$Assuming $h_1 
eq h_2$,we get $H = h_1 + h_2$.This implies that the ratio $h_1/h_2$ is not a fixed constant but depends on the specific values of $h_1, h_2$ and $H$ such that their sum is $H$. Thus,the ratio depends on $H$.

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