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Egg drop from building

  1. Jan 27, 2008 #1

    You are on the roof of a physics building, which is 46.0 m above the ground. Your physics professor, who is 1.80 m tall, is walking alongside the building at a constant rate of 1.20 m/s. If you wish to drop an egg on your professor's head, how far from the building should the professor be when you release the egg?

    2. Relevant equations

    3. The attempt at a solution

    I took 46.0 and subtracted 1.80 for the height of the professor. That gave me 44.2 m. I then divided that by 9.80 m/s^2 for acceleration due to gravity. That gave me 4.51 s. I then divided that by 1.20 m/s for the velocity of the professor. This gave me an answer of 3.76 m. I typed this in and the masteringphysics said not quite, that I may have a slight error in calculuation or used the wrong number of significant figures. I recalculated several times and kept getting 3.76. Any ideas?
  2. jcsd
  3. Jan 27, 2008 #2


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    44.2m divided by 9.80 m/s^2 yields 4.51 seconds squared.
    Careful with those units. You're solving an equation not cooking up a spell in a cauldron. Work it out carefully from the actual equations, e.g.
    [tex] x(t) = \frac{1}{2}a t^2 + v_0 t + x_0[/tex]
    You're solving for [itex]t[/itex] in this equation with [itex]v_0 = 0[/itex] and with [itex] x-x_0 = 44.2 m[/itex].
    The answer is then not [itex] t = 4.51[/itex] seconds.
  4. Jan 27, 2008 #3
    Using your formula, I got 3 seconds for the drop and the professor should be 2.5 meters. is that more accurate?
    Last edited: Jan 27, 2008
  5. Jan 28, 2008 #4


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    It isn't a matter of accuracy but of correctness. You should have used this formula from the start. Just because you can often finagle out a multiple choice answer by doing alchemy on the numbers given in the problem doesn't mean you're learning anything. You MUST work the problem using the physics of the circumstance. This formula should be in your text. You should have invoked it immediately.

    As to whether this is the correct answer, you'll have to submit it and see if the computer accepts it as correct, but the formula represents the position of a constantly accelerated object given its initial position and velocity. This is the case for your water balloon. You can as easily apply it to the professor with a=0, and make sure your figurin' is correct with respect to how far he moves in the time you've determined.

    Your original approach however makes me despair for our young students.
  6. Jan 28, 2008 #5
    It's not a multiple choice question. You have no idea what I have been going through the last few weeks, so don't judge me. Watching someone you care about take their last breaths makes it a little bit more difficult to focus on which formulas work for which situations. If you want to point me in the right direction, fine. But if you're going to condescend, I'll get help elsewhere.
    Last edited: Jan 28, 2008
  7. Jan 28, 2008 #6


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    I beg your pardon. You're correct I shouldn't have made assumptions, nor expressed them.
    You have my deepest apology.

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