As a body falls its acceleration is given by `a(h) = 8/h` where h is the height. How long does the body take to fall from 120 m?

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txmedteach | High School Teacher | (Level 3) Associate Educator

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Alright, I'm going to start off by saying that this looks NOTHING like any real physics equation. Now that I got that off my chest, let's answer the question.

`a(h) = -8/h`

We're going to presume that acceleration is negative because it's downward, and that it's in the standard units of `m/s^2` (Meters per second squared). This equation also says that as h approaches 0, a(h) will become negative infinity (suspicious). At least it puts a limit on the time of our fall!

Now, the problem takes a dive. You should recognize acceleration is the second derivative of height with respect to time:

`a(h) = (d^2h)/(dt)^2`


`(d^2h)/(dt)^2 = -8/h`

Unfortunately, this is definitely a nonlinear second-order differential equation. Try as I might, I could not find a good way to solve this, except numerically. The numerical solution is found by calculating the initial acceleration, giving you your velocity (depending on the time step, I used 0.1 seconds), which will then give you your new height when multiplied by the time step. So, here are the numerical steps (given a step n and time step (`Deltat`) 0.1):

`a_n = -8/(h_(n-1))`

`v_n = v_(n-1) + a_n*Deltat`

`h_n = h_(n-1) + v_n*Deltat`

Now, these steps have initial conditions:

`a_0 = 0`

`v_0 = 0`

`h_0 = 120`

Now, given all of that (and a spreadsheet), I found a time of 53.2 seconds.

Now, some people will simply integrate with respect to h initially. Let me tell you that is absolutely incorrect. Acceleration is a derivative with respect to time, and if you try to differentiate/integrate with respect to distance you get something more analagous to energy than velocity (with an unknown proportionality constant!).

SO, this is the only way to do it unless you can solve nonlinear second order differential equations.

Hope that helps!

Side note: this is how your calculator solves D.E.'s (if it can).