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Stuck on Free-Fall Acceleration problem

  1. Jan 23, 2006 #1
    I keep trying to get this right but I keep falling short. Here's the problem:

    A hoodlum throws a stone vertically downward with an initial speed of 19.0 m/s from the roof of a building, 39.0 m above the ground. (a) How long does it take the stone to reach the ground? (b) What is the speed of the stone at impact?

    My professor has given us a number of formulas to use for motion problems and they are as follows:

    Vf = Final Velocity, Vi = Initial Velocity, a = acceleration, t = time, d = displacement

    First Formula: Vf^2 = Vi^2 + 2ad

    Second Formula: d = Vi(t) + (a/2)(t^2)

    Third Formula: Vf = Vi + at

    Which formula could I use to solve this problem? For part a based on what my textbook indicates it'd seem that the second formula will do it, but when I plug in the data that's given I end up with:

    39m = 19t - 4.9t^2

    But as you can see this doesn't give me a single variable. (Well it does, t, but I end up with a quadratic equation).

    Do I just solve the quadratic or am I doing something wrong? And if I'm doing something wrong, am I using the right formula of the 3 or do I need to use a different one? Please help! Thanks!
  2. jcsd
  3. Jan 23, 2006 #2
    You are using the correct formula, but all the terms should be of the same sign, positive or negative, since the displacement, initial velocity, and acceleration all have the same direction - downward.
  4. Jan 23, 2006 #3
    Rewriting then...

    -39m = -19t - 4.9t^2?

    But no matter how I setup the signs I'm stuck with a quadratic. Do I just solve like a quadratic then?
  5. Jan 23, 2006 #4

    What is wrong with having a quadratic? Solve for t. You'll get two answers, one positive, one negative. Obviously, it can't be the negative one (since that would be before the stone was thrown). So the positive one is your answer. However, your quadratic has the wrong signs. The correct equation should be:


    This is because d is the distance the stone travels, not its initial height. So if it goes 39m down, it moves -39m. If its velocity is 19 m/s down, that -19m/s. If its acceleration is 9.8 m/s^2 down, thats -9.8 m/s^2.
  6. Jan 23, 2006 #5
    Ahh thanks. Well as long as a quadratic is fine I'm good to go. Thanks again!
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