APEX Calculus, p.194, Ex. 32

The distance, in feet, a stone drops in \(t\) seconds is given by \(d(t) = 16t^{2}\). The depth of a hole is to be approximated by dropping a rock and listening for it to hit the bottom. What is the propagated error if the time measurement is accurate to 2/10ths of a second and the measured time is:

  1. 2 seconds?
  2. 5 seconds?

\[D=16t^{2}\] \[dD = 32t dt\]

Given that the accuracy of the time measurement is \(\pm 0.2s\), the propagated error is approximately:

\[dD = 32t*0.2s\]

  1. For a fall of 2 seconds that equates to:

\[dD(2) = 32*2*(\pm0.2)\] \[dD(2) = \pm12.8ft\] Considering the true fall distance at 2 seconds is \(16*2*2=64ft\), 12.8ft of error if significnant.

  1. For a fall of 5 seconds that equates to:

\[dD(5) = 32*5*(\pm0.2)\] \[dD(5) = \pm32ft\] At an actual fall distance of \(16*5*5=400\) this 32ft error doesnโ€™t seem all that bad.