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    A simple physics problem

    By jim On 29 October 2003 · 5 Comments · In Cool Geek Tricks

    A simple physics question was asked yesterday at work, but it generated a lot of debate: “A guy’s on top of a 20 foot ladder when it falls over. How fast was he going when he hit the ground?”

    For simplicity, ignore issues such as the foot of the ladder moving, air resistance, seats, humidity, etc. Also assume the guy doesn’t fall off the ladder, but follows it down. (This is as realistic as a spherical, frictionless, non-conducting bovine.)

    If our friend was just dropped from 20′ above the ground, we could calculate the time to fall as:

    Distance = Average velocity * time

    Which works out to:

    20 feet = 1/2 * 32 feet/s2 * t2

    or, solving for time:

    =Time = Distance / Average velocity

    which is:

    Time = sqrt [ 2 * (20/32) ]
    T = 1.12 seconds

    Velocity at impact would be:

    Velocity = Acceleration * time
    = 32 ft/s2 * 1.12 s
    = 35.8 ft/s, or 24.3 miles per hour

    All this is straightforward. The controversy came in because our friend is traversing an arc of 31.4 feet through space. What changes?

    Answer: nothing.
    Using the calculations from Amusement Park Physics, this is just a pendulum. We know the potential energy at the top, and the kinetic energy at the bottom. With the same assumptions as above, we can solve for velocity.

    Kinetic Energy, Ek = 1/2 * mass * velocity2
    Potential Energy, Ep = mass * G * height

    Ek = Ep
    1/2 * m * v 2 = m * G * h

    1/2 * m * v 2
    = m * G * h

    = sqrt ( 2 * G * h)
    = sqrt ( 2 * 32 feet / s2 * 20 feet)
    = 35.8 ft/s, or 24.3 miles per hour

    Practically speaking, our friend broke his leg.

    There were numerous creative ways to solve the problem the wrong way. And as an engineer who hasn’t taken physics in nearly 20 years, I pursued a couple of these before realizing this is the correct answer.

    A more interesting extension is the reality where the guy falls off the ladder and becomes a projectile. This is too much to think through right now.
     

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    5 Responses to A simple physics problem

    1. Director Mitch says:
      29 October 2003 at 11:24

      When reading this I thought of it as a projectile problem, remembering that x and y movement are independent of one another. The downward velocity he hits the ground is just a calculation of the y vector, which you have above. However, if he didn’t ride the ladder down and fell off in an arc, you could argue that there will be an x component (however small) to the velocity, which would create a total velocity vector angled INTO the ground which will be a combination of his downward y vector and his sideways x vector, which you would solve as the hypotnuse of the right angle vectors, meaning his total velocity will be greater than just the downward pull of gravity:

      —-> Sideways Velocity Vector
      |\
      | \
      | \
      | \
      | _\| Total Velocity Vector
      \/
      Downward Velocity Vector

      But if he actually rides the ladder down in the pendulum method you describe, he probably wouldn’t have any x velocity, so that part becomes zero and you have just the downward y velocity.

      The REAL question is: who is liable? Obviously since Walmart sold the ladder in question (and has the deepest pockets) they should pay through the nose for our friend’s leg… :-)

      Reply
    2. carson says:
      29 October 2003 at 22:12

      The assumptions greatly simplify the problem and are not reality.

      When I originally tried the problem, I used your method and drew a bunch of small triangles with sides x = v * sin T and y = v * cos T.

      I don’t know where I’d expect him to depart the ladder on his freeform trajectory, but once he does, his X velocity remains constant until he hits the ground. From that departure point, the trajectory looks like the hyperbola: Y = – x^2

      Did I mention this is more physics than I’ve done in a long time?

      Reply
    3. Director Mitch says:
      30 October 2003 at 11:10

      Yeah, my assumption was that he jumped away from the ladder, giving him x momentum because he was, uh, trying to reach for a ledge…yeah, that’s it.

      > Did I mention this is more physics than I’ve
      > done in a long time?

      I keep telling people that I spent 10s of 1000s of dollars getting an electrical engineering degree and don’t use a damn thing I learned. That might be simplifying it a little bit since I do have to understand how my high-tech components work and having fundamental understanding of electricity, semiconductors, etc. is helpful, but it’s not like I had to go to college to learn all that….I learned semiconductor manufacturing at a course I took while at my first company, I didn’t learn a damn thing about that in school.

      (okay, the argument is college makes you a better rounded, educated individual, better to participate in society…)

      Reply
    4. carson says:
      30 October 2003 at 13:30

      > getting an electrical engineering degree and don’t use a damn thing I learned.

      This would be a great topic for a blog entry this weekend. Much as you learned semiconductor manufacturing from your first employer, so too did I learn database design.

      I’m a little saddened to think about how little I’m using my MBA right now… Some days I think I would have been better off taking a course in cat herding instead. But then I wouldn’t enjoy your discourses on economics…

      Reply
    5. anna says:
      16 June 2007 at 11:17

      time=sqrt[2*(20/32)]t=1.12 seconds

      Reply

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