In a cylindrical water tank, there are two small holes $$A$$ and $$B$$ on the wall at a depth of $$h_1$$, from the surface of water and at a height of $$h_2$$ from the bottom of water tank. Surface of water is at height $$h_2$$ from the bottom of water tank. Surface of water is at height $$H$$ from the bottom of water tank. Water coming out from both holes strikes the ground at the same point $$S$$. Find the ratio of $$h_1$$ and $$h_2$$.

The horizontal range of water exiting a hole in a tank is given by $$R = 2\sqrt{h_{\text{depth}} \cdot h_{\text{height}}}$$.

Given: For hole A: $$\text{depth} = h_1,\quad \text{height} = H - h_1$$

For hole B: $$\text{height} = h_2,\quad \text{depth} = H - h_2$$

Equating the horizontal ranges ($$R_A = R_B$$):

$$2\sqrt{h_1(H - h_1)} = 2\sqrt{(H - h_2)h_2}$$

$$h_1H - h_1^2 = h_2H - h_2^2$$

$$H(h_1 - h_2) = h_1^2 - h_2^2$$

$$H(h_1 - h_2) = (h_1 - h_2)(h_1 + h_2)$$

Since the two holes are at different locations ($$h_1 \neq h_2$$):

$$H = h_1 + h_2 \implies h_1 = H - h_2$$

$$\frac{h_1}{h_2} = \frac{H - h_2}{h_2} = \frac{H}{h_2} - 1$$