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Old Sep 1st 2009, 11:40 AM   #1
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Free Fall

Water drips from the nozzle of a shower onto the floor 200 cm below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. Find the locations (above the floor in cm) of the second and third drops when the first strikes the floor.
I think I understand how to do this question, but the system won't accept my answer.

I converted 200 cm to 2 m beforeI did anything else.

I started by finding the final velocity of the first drop using dx = (v^2-iV^2)/(2a). I then used the velocity to find the total time of the first drop using dV = at. I got the time to be .23 seconds.

To find the position of the second and third drops i set up proportions for the times. For example the second drop was at .1533 seconds of its fall. I plugged that number into dx = iVt + 1/2(at^2) and got 75.11 (converted back to cm). I then subtracted that from 200 and got 124.883.

Does anyone see where I went wrong?
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Old Sep 1st 2009, 12:01 PM   #2
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You certainly make things hard for yourself.

Try dividing the four drops by 3 equal time pitches over the total distance.

Last edited by Paul46; Sep 1st 2009 at 12:52 PM.
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Old Sep 1st 2009, 03:38 PM   #3
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I thought about that, but then I figured that the distance wasn't proportional since velocity was constantly increasing.

Anyway, after redoing it again I got the right answer. I had some small computation error.
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Old Sep 1st 2009, 11:06 PM   #4
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Well, if you find the time of the total distance and split the total time by 3 equal pitches your find the answer.

2m / (1/2) x acceleration = Time^2

drop 1 = 0.6388765 s = (1/2)at^2 = 2m

If you had divided the time by 3 equal time pitches you would of found the answers.

Drop 2 = 200 - 88.88 (2 d.p) = 111.12 cm from floor

drop 3 = 200 - 22.22 (2 d.p) = 177.78 cm from floor
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