Consider the figure of a fountain with four holes given below : Which one of the holes in the fountain will throw the water farthest ?

Water jets from holes are driven by hydrostatic pressure, which increases with depth; this causes higher exit speed for holes deeper below the free surface [1]. By Torricelli’s law the horizontal exit speed from a hole at depth h below the free surface is v = sqrt(2gh) [2]. If the free-surface height above ground is H, a hole at depth h has horizontal range R = v·t = sqrt(2gh)·sqrt(2(H−h)/g) = 2·sqrt[h(H−h)]. This product is maximum when h = H/2, i.e. for the hole nearest mid-depth. With holes numbered 1 (top) to 4 (bottom) and equally spaced, hole 2 lies closest to mid-depth, so it gives the farthest jet [2].

Sources

1. [1] Science ,Class VIII . NCERT(Revised ed 2025) > Chapter 6: Pressure, Winds, Storms, and Cyclones > Activity 6.2: Let us find out > p. 85
2. [2] https://en.wikipedia.org/wiki/Torricelli%27s_law