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I'm sure a simple question.

I have a video of me jumping off a cliff into a river. I want to calculate how high it is. I know my weight, acceleration due to gravity of course, and I can get the time it took to hit the water.

Can I calculate (albeit roughly) how high the cliff was?

Cheers,

Tim.

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I think it is important to appreciate that your weight plays absolutely no role for this question (as evident from the answer below). –  Daniel Grumiller Feb 28 '11 at 10:15
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@Daniel: ... except if you take air resistance into account, in which case you need other parameters too. –  Frédéric Grosshans Feb 28 '11 at 16:03
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2 Answers 2

up vote 4 down vote accepted

If I remember correctly, the distance should be $y=\frac{1}{2} g t^2$. This is for if you simply fell off the cliff. Now if you jumped off the cliff, it will be different because you may have some initial velocity in the y-direction. Also, of course, air friction is not being accounted for.

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Cheers mate! Plugging the numbers in, it's sounding just about right by my judgment, allowing for these other factors you mention...(about 10m, as I thought - good fun). –  user2313 Feb 28 '11 at 5:40
    
If you want to take air resistance into account, look at this wikipedia page en.wikipedia.org/wiki/… –  Frédéric Grosshans Feb 28 '11 at 10:45
    
The initial velocity in the y direction is irrelevant, no ? The vertical velocity (z) matters. –  Frédéric Grosshans Feb 28 '11 at 16:20
    
@Frederik I think my y is your z –  BeauGeste Feb 28 '11 at 19:04
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Ignoring air resistance is a good approximation in this case:

  • your resolution of the actual time is probably pretty coarse so there's (EDIT TO ADD: no) need to make other parts of your calc significantly more accurate than your least-accurate portion.
  • your speed was quite a a bit lower than your terminal velocity, therefore the drag would have been a very small force relative to gravity's.

To confirm this, do the calculation with air resistance set to zero, and then see if your initial estimate of time and impact velocity are in the "air resistance is negligible" regime, or if it could have been significant.

So: with no air resistance d=1/2*a*t^2, and in ballpark numbers d=10m and a=10m/s^2 -- so t=1.4seconds; this should match your measured time, if not work backwards from your measured time to get "d" in this no-air-resistance scenario.

Work out your highest speed i.e. just prior to impact: v=a*t. With the above numbers we get v=14m/s

Now compare this speed to the speed at which air resistance matches your weight. From skydivers we know that a human body has a terminal velocity of about 120mph in a horizontal position and about 180mph in a diving position. Going with the lower one, 120mph is about 180ft/sec which is about 60m/s.

So you reached about 1/4 of terminal velocity (14ms / 60m/s). Air drag scales with the square of velocity so the greatest air drag force you experienced (just prior to hitting the water) was (1/4)^2 or 1/16th of your weight, which is about a 6% error. Averaged over your fall it would have been much less than 2%, and since your timing probably isn't to 1%-2% precision, it is therefore a sound judgement to neglect air resistance.

EDIT TO ADD: d=1/2*a*t^2 is the standard formula for the distance covered by an object under constant acceleration, starting from rest. It can be obtained by integrating the formula v=a*t which is the standard formula for velocity attained under constant acceleration, starting from rest.

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Thanks Daniel for that thoroughly comprehensive answer. Could you help me once more by briefly explaining how you get 1/2*a*t^2 as the forumula here? I'm sure you're right - I just need intellectual satisfaction! (It's been a while...) –  user2313 Feb 28 '11 at 22:15
    
@Tim: note that the speed $v = at$ is not constant but instead increases linearly with time. The same can be said about your position $x = vt$. So if you just plug in the $v$ you'll get $x = at^2$. Now, this gives a good order estimate but it's not quite right because the average velocity during your flight wasn't $v$ but $v/2$ (you started at zero and ended at $v$). –  Marek Feb 28 '11 at 22:42
    
Thanks, Marek - I found that a good and useful explanation. And thank you again to everybody. –  user2313 Mar 1 '11 at 0:20
    
+1 for actually evaluating the importance of air resistance –  Frédéric Grosshans Mar 1 '11 at 14:26
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