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If I wrote freshman-year kinematics exams…

May 15, 2011

…I would definitely include the following problem:

You live in the dorms and your upstairs neighbor, LeBrian Skinner, is a serious basketball player.  He is about to declare for the NBA draft, but he fears that his merely average height will put him at a disadvantage.  To compensate for his relative shortness, LeBrian decides that he needs to have a vertical jump of at least 36 inches.

In the evening you can hear LeBrian practicing his vertical leap, since he lives directly above you: you hear a loud creak when he first jumps followed by a loud thump when he lands again.  You use a stopwatch to time the interval between the moment he first leaves the floor and the moment when he lands again.  You measure this interval as 0.8 seconds.

Assuming that LeBrian lands with his legs fully extended (in the same position as when he leaves the floor), how high is he jumping?  Is it enough?

For those who are curious, the solution is after the page break.

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Solution:

(sorry for the poor image quality).

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Of course, in the real world the answer “31 inches” isn’t enough.  You need to have a sense of how accurate that answer is (are you really sure that LeBrian has no chance of getting drafted?)  I personally think that all freshman-level physics classes should be required to teach their students enough to answer this second part of the question:

Suppose you only measured the time interval between takeoff and landing of LeBrian’s jump with an accuracy \pm 0.1 seconds.  What is the uncertainty in your calculation of the height of the jump?

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The solution to this second part goes like this:

Let \Delta t = 0.1 \textrm{ s} be the uncertainty in the timing.  Call \Delta h the uncertainty in the height of the jump.

In part 1 of the question, we derived h = g (t/2)^2/2, where t \approx 0.8 \textrm{ s} is the full duration of the jump.

The uncertainty in the jump height \Delta h satisfies \Delta h = (dh/dt) \Delta t.

The derivative dh/dt = g t/4.

So \Delta h = g t/4 \times \Delta t = (9.81 \textrm{ m/s}^2) \times (0.8 \textrm{ s}) \times (0.1 \textrm{ s})/4 = 0.196 \textrm{ m} = 7.7 \text{ inches}.

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So in the end, the answer is that LeBrian’s vertical jump is something like 31 \pm 8 inches.

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By the way, I’m not sure what this problem says about me.  Apparently I’m not arrogant enough to give even my fictional alter ego a great chance of making it to the pro’s, but I’m arrogant enough to say that it’s within uncertainty.

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7 Comments leave one →
  1. Repton permalink
    May 16, 2011 1:46 am

    The science museum in Canberra had (last time I visited, about 10 years ago) a doohickey like that: you stand on a plate, jump into the air, and land on the same plate, and it tells you how high you jumped.

    At the time I had the thought to try to fool it by tucking my legs, but later I realised that it was just timing my jump and calculating.

    • Paul Muller permalink
      May 16, 2011 10:21 am

      There are now mobile phone apps that do the same thing. You hold the phone, it senses your takeoff and landing and times the interval, then it does the math and teels you how high you jumped.

    • gravityandlevity permalink*
      May 16, 2011 10:37 am

      Wow, I am apparently significantly behind the times.

  2. May 17, 2011 11:49 pm

    I see several flaws…first, there is the assumption that his “average HEIGHT'(relative) is a disadvantage. With a variance of +-.01 … nevermind…he was short!

  3. L.B. Jeffries permalink
    May 18, 2011 8:22 pm

    Cool post! Thanks.

  4. atomsinmotion permalink
    June 1, 2011 2:35 pm

    I once used the inverse of this technique to estimate my reaction time.

    I was putting clothes in a top-loading clothes-washer and some water spilled out of the machine.

    I noticed that I started to move (to avoid getting wet) just as the water hit the floor and thereby estimated my reaction time to be about 0.2 secs.

    Coincidentally a friend of mine had a knock to the head at about the same time and was having his reactions tested to check for concussion.

    It turned out that my estimated time was pretty close to the normal human value as reported by him.

  5. jasman permalink
    December 23, 2011 12:39 pm

    I like it, but ur coordinate system is drawn incorrectly i think. To be consistent with how are you solve the problem I think you need to shift the time coordinate it to the top of the parabolic arc, so y(t, 0.4) equals 0 when v(t 0.4) equals 0. Anyway, nice.

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