- #1
Josh0768
- 53
- 6
Suppose I have a perfectly circular pool which is four meters in radius, two meters in depth, and filled with water. Say I drop a steel ball with a radius of five centimeters into the middle of the pool from a height of five meters above the water's surface. After three seconds, what will be the approximate height (relative to the surface of the water before it was perturbed) be of a point on the water's surface that is two meters horizontal from the center of the pool? Ignore any contribution to the volume of water made by the steel ball.
*Please note: original specifications were a pool depth of three meters, radius of five meters, and time elapsed of six seconds. These parameters have been changed to those you see now.*
*Please note: original specifications were a pool depth of three meters, radius of five meters, and time elapsed of six seconds. These parameters have been changed to those you see now.*
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