1.9: Exercises

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Before you go on, you might want to work on the following exercises.

Exercise \(\PageIndex{1}\)

You might have heard that a penny dropped from the top of the Empire State Building would be going so fast when it hit the pavement that it would be embedded in the concrete or that if it hit a person it would break their skull.

We can test this myth by making and analyzing a model. To get started, we’ll assume that the effect of air resistance is small. This will turn out to be a bad assumption, but bear with me.

If air resistance is negligible, the primary force acting on the penny is gravity, which causes the penny to accelerate downward.

If the initial velocity is 0, the velocity after \(t\) seconds is \(a t\), and the distance the penny has dropped is \[h = a t^2 / 2\notag\] Using algebra, we can solve for \(t\): \[t = \sqrt{ 2 h / a}\notag\] Plugging in the acceleration of gravity, \(a=9.8\,\textrm{m/s}^{2}\), and the height of the Empire State Building, \(h=381\,\textrm{m}\), we get \(t = 8.8\,\textrm{s}\). Then, computing \(v = a t\) we get a velocity on impact of \(86\,\textrm{m/s}\), which is about 190 miles per hour. That sounds like it could hurt.

Use MATLAB to perform these computations, and check that you get the same result.

Exercise \(\PageIndex{2}\)

The result in the previous exercise is not accurate because it ignores air resistance. In reality, once the penny gets to about \(18\,\textrm{m/s}\), the upward force of air resistance equals the downward force of gravity, so the penny stops accelerating. After that, it doesn’t matter how far the penny falls; it hits the sidewalk at about \(18\,\textrm{m/s}\), much less than \(86\,\textrm{m/s}\).

As an exercise, compute the time it takes for the penny to reach the sidewalk if we assume that it accelerates with constant acceleration \(a=9.8\,\textrm{m/s}^{2}\) until it reaches terminal velocity, then falls with constant velocity until it hits the sidewalk.

The result you get is not exact, but it’s a pretty good approximation.