1. 2010
    Apr
    25

    Calculating terminal speed

    In the 22,000 Foot Fall episode of Mythbusters, Adam did a calculation of how long it should take for a falling person to reach terminal speed. It occurred to me that there’s a wrong (but simple) way and a right (but complicated) way to do this calculation — I wonder which one was used on the show to come up with 487 feet?

    First, the simple way. From numerous tests in previous episodes, the Mythbusters know that the terminal speed of a person falling through air is about 120 miles per hour. Based on that, you could try to figure out the fall height it takes to achieve that speed using the formula

    $$h_0 = \frac{v_T^2}{2g}$$

    (It’ll become clear later on why I’m calling the height \(h_0\)). This formula comes from the kinematic equation \(v^2 = v_0^2 + 2ad\) with initial velocity \(v_0 = 0\), final velocity \(v = v_T = \unit{120}{\mileperhour}\), and acceleration \(a = g = \unit{9.8}{\frac{\meter}{\second^2}}\). Plugging in the numbers and calculating gives

    $$h_0 = \unit{147}{\meter} = \unit{482}{\foot}$$

    That’s reasonably close to what Adam got on the show, so I’m guessing this is the method …