00:01
So here we have to define a volumetric flow rate.
00:04
So this is going to be the cross -sectional area of whatever section of the bucket, pipe hole, whatever it may be, times the velocity through that section.
00:15
And then due to the conservation of energy, where all potential energy is converted into kinetic energy, the velocity is going to equal the square root of 2gy, which is simply defines the velocity.
00:31
After a free fall of distance y.
00:42
So after this we can simply substitute in.
00:46
Q is simply going to be a times the square root of 2gy.
00:50
This is going to be q squared equals a squared 2gy.
00:56
And then we have to isolate this y.
00:59
So y is simply going to be equal to q squared divided by a squared times 2g.
01:07
At this point, we can simply substitute in.
01:11
So, y would be equal to that volumetric flow rate, which is going to be 2 .4 times 10 to the negative 4th...