Testing Galileo

Congratulations, you figured out how far a rock will fall each second assuming there's no air resistance (just like Galileo did):

• 1 sec = 16 ft
• 2 sec = 64 ft
• 3 sec = 144 ft
• 4 sec = 256 ft
• 5 sec = 400 ft
• 6 sec = 576 ft
• 7 sec = 784 ft
• 8 sec = 1024 ft

But how much difference does the air make? While soaking in your hot-tub next to your waterfall (see below), you decide to graph the velocity vs time for three different sized rocks up to 20 seconds. First graph the velocity in the absence of air (using 32 ft/s² for gravity). Next, add the velocity for rocks with 1 cm, 3 cm and 10 cm diameters. Use Wolfram Alpha to figure out the velocity by pasting in this equation:

v=sqrt(x)*tanh(sqrt(1/x)*9.8*t)*3.28

which gives the velocity in ft/s for t in seconds. The variable x is equal to 327*m/a where m is the mass in grams and a is the cross-sectional area of the rock in cm² (assume the rocks are spheres). Rocks weigh about 2.7 g/cm³. Once you have values of x for each size rock, you can use Wolfram Alpha to show the graph of velocity vs time. For example, for x =1000 you can enter this into Wolfram Alpha:

v=sqrt(x)*tanh(sqrt(1/x)*9.8*t)*3.28 where x=1000

To get the velocities after the first five seconds, you can use this:

v=sqrt(x)*tanh(sqrt(1/x)*9.8*t)*3.28 where x=1000, t={0,1,2,3,4,5}

How quickly do the different rocks slow down? Do you need to modify your table for the air resistance? Do you need different tables for different size rocks?

Calculate a new table for each size rock using this:

integrate sqrt(1000)*tanh(sqrt(1/1000)*9.8*t)*3.28 dt from 0 to 1

but replace '1000' with the value of x for each rock and the last '1' with the number of seconds the rock falls.