4. At depth Y in cm, the cross-sectional area of a cylinder is 5*Y+10 times as great as the cross-sectional area of the hole. Inital depth is 50 cm. Give your best approximation of the water depth in the first three 10-second intervals. Exit Velocity is Square Root(1960*Y).

Let Y = Depth
Let Z = Exit Velocity
Let X = Cross Sectional Area of the Hole

Cross Sectional Area of the Cylinder:
5Y+10(X)
5 * 50 +10 ( Initial Depth = 50)
260 * X cm^2

Z = SQRT(1960*Y) cm/sec

Volume of the Cylinder:
50 cm * (260 * X) cm^2
13,000 X (cm^3)

Multiply Z by the interval of time. Subtract this from 13,000 X (cm^3). This will give you how much water is left in the cylinder during the time period. Divide this by 260 X cm. This will give you remaining Depth in cm. Repeat using new depths.
Assuming X=1
  
Depth
TimeDepth
0 sec50 cm
10 sec37.97 cm
20 sec27.48 cm
30 sec18.5 cm