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Chapter 12
Terminal Velocity
Suppose you drop a penny from the top of the Empire State Building, and you want to know how fast it is going when it hits the ground. Let y represent the height, in feet, of the penny t seconds after you drop it. The Empire State Building is 1250 feet tall, so y = 1250 when t = 0. Since you drop it and don’t throw it down, the initial velocity is 0 feet/sec, namely, y′^ = 0 when t = 0. Finally, gravity will be accelerating the penny at -32 feet/sec^2. So y will satisfy:
- y = 1250 when t = 0
- y′^ = 0 when t = 0
- y′′^ = −32 for all t.
Since y′^ is an antiderivative of y′′^ = −32, we get that y′^ = − 32 t + C for some C. When t = 0, we get both y′^ = 0, from (2) above, and y′^ = − 32 · 0 + C = C, from our formula. So C = 0, and y′^ = − 32 t. Next, since y is an antiderivative of y′^ = − 32 t, we get y = − 16 t^2 + C 0 for some C 0. When t = 0, we get both y = 1250, from (1) above, and y = − 16 · 02 + C 0 = C 0 , from our formula. So y = − 16 t^2 + 1250. To find the speed when the penny hits the ground, note that it hits the ground when y = 0, i.e., − 16 t^2 + 1250 = 0. This gives us t = 8. seconds. In t = 8.8 seconds, the velocity will be y′^ = − 32 · 8.8 = −281. feet/second, which gives us a speed of about 192 miles/hour. (Recall, the speed doesn’t have the minus sign.)
CHAPTER 12. TERMINAL VELOCITY 2
However, we left off some information. The acceleration due to grav- ity changes depending on the height of the penny, but the amount that changes is negligible. We also ignored air resistence, which is not negli- gible. (There are plenty of other things we left off, such as the effect of being next to a building has on the air, but we will only consider the air resistence.) Air resistence is tricky to deal with, particularly for something like a tumbling penny, where the cross-sectional area is constantly changing. In general, for small velocities, air resistence is about proportional to velocity, and for larger velocities, air resistence is about proportional to velocity squared. For our approximation, we will assume that the acceleration due to air resistence is proportional to velocity squared, so if y is the height, then the acceleration due to wind resistence will be k(y′)^2. Here, k will be positive, since the force of air resistence will be up. For a penny, we’ll estimate k = 0.025. So when the penny is dropped, gravity will push down and air will push up, so the acceleration will satisfy y′′^ = − 32 + k(y′)^2 , or y′′^ = − 32 + 0.025(y′)^2
with the conditions that when t = 0, we also have y = 1250 and y′^ = 0. This can be solved with great difficulty, but the velocity will end up being
y′^ = 35.
1 − e1.78885t 1 + e1.78885t
feet/sec
As times goes by, the velocity will approach −35.7771 feet/sec, or about −24.4 miles/hour. This is called the terminal velocity. Note that we can find the terminal velocity without actually solving the differential equation. From
y′′^ = − 32 + 0.025(y′)^2
we can note that terminal velocity is when the object does not accelerate, namely y′′^ = 0. This gives us
− 32 + 0.025(y′)^2 = 0
