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A rock is THROWN downward. What was the initial height?

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data

    A rock is thrown downward from an unknown height above the ground with an initial speed
    of 27 m/s. It strikes the ground 7.1 s later. Determine the initial height of the rock above the ground. The acceleration of gravity is 9.8 m/s2. Answer in units of m.

    2. Relevant equations

    [itex]\Delta[/itex]x = vit + 1/2 at2

    3. The attempt at a solution

    I'm not sure if this equation is correct, first of all, but it has all of the variables given in the problem. I don't know how to deal with the fact that it's thrown down, but I thought the vi would take care of it.

    We have acceleration, initial velocity, and time, and the equation solves for displacement. What am I doing wrong? How should this problem be set up differently and why?

    Thank you very much.
     
  2. jcsd
  3. Sep 28, 2011 #2

    Doc Al

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    Staff: Mentor

    It will.

    Looks good so far. What's preventing you from solving it?

    Be careful with signs.
     
  4. Sep 28, 2011 #3
    We're using a web application that checks the answers for us, and I got it wrong when I entered what I got. I'll try again with the same method, though. I may have messed up the signs, but I also feel like this problem takes more than just plugging in to an equation.
     
  5. Sep 28, 2011 #4
    A lot of times web applications have to have a specific format for both the values (including proper use of significant figures and rounding) as well as the units.
     
  6. Sep 28, 2011 #5

    Doc Al

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    Staff: Mentor

    In order to know where you went wrong, we need to know what values you plugged in and what was your result.

    (Regardless of the sign convention you use in solving the problem, the answer that they want will be positive.)
     
  7. Sep 28, 2011 #6
    I solved this again and got -55.309. This is not correct, and neither is its absolute value.

    I'm confident in my 4th grade order of operations ability, so the problem probably lies in how I set this up. Here is what I solved:

    [itex]\Delta[/itex]x = 27(7.1) + 1/2(-9.8)(7.1)2
    [itex]\Delta[/itex]x = -55.309 Incorrect
     
  8. Sep 28, 2011 #7

    Doc Al

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    Staff: Mentor

    Since the rock is thrown downward, what must be the sign of its initial velocity?
     
  9. Sep 28, 2011 #8
    Thought about that but hesitated to try it. The absolute value of the solution was correct! Thank you very much, Doc Al. I'm still getting used to watching my signs.
     
  10. Sep 29, 2011 #9
    That's a sticking point for a lot of folks when they first start getting into the equations of motion. Just remember that g = -9.8m/s2 only if you define the upward direction to be positive, so in your example, since you throw it downward, the initial velocity is negative.
     
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