Projectile analysis of stream, obtaining cross-sectional area from surface velocity

If the stream of water exiting a container in an initially horizontal stream has a horizontal range of 30 cm from an altitude of 90 cm, and if at this instant the water surface is descending at 1.5 millimeters per second, then what is the ratio of the cross-sectional area of the container to the area of the exit hole?

We find the velocity of the horizontal stream by a projectile analysis.  The ratio of areas is the inverse ratio of velocities.

A range of 30 cm for a fall of 90 cm implies time of fall `dt = sqrt( 2 * .9 m / (9.8 m/s^2) ) = .43 sec, approx..

A horizontal displacement of 30 cm at constant velocity in .43 sec implies velocity 30 cm / (.43 sec) = 70 cm/s, approx..

The ratio of exit velocities is therefore 70 cm/s / (.15 cm/s) = 470.

Thus the cross-sectional area of the container is about 470 times that of the hole.