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Ball thrown straight up

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1 second later. You may ignore air resistance

    What must the height of the building be for both balls to reach the ground at the same time if vo is 6.0m/s?

    2. Relevant equations

    y=yo+vo*t+1/2*a*t^(2)

    t2=t1-1seconds


    3. The attempt at a solution

    I chose positive going down

    ball 2:

    y=yo+vo*t +1/2a*t^(2)

    y=4.905m/s^(2)*t2^(2)
    y=4.905m/s^(2)*(t1-1s)^(2)

    y=4.905m/s^(2)t1^(2)-9.81m/s*t1+4.905m

    ball 1

    y=yo+vo*t +1/2a*t^(2)

    y=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

    Set the final equations equal to each other:
    y=y

    4.905m/s^(2)t1^(2)-9.81m/s*t1+4.905m=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

    4.905m=3.81m/s*t1

    t1=1.2874 seconds

    plug t1 back into equation for ball 1

    I get 0.4051 m but the answer is 0.411 m

    What did I do wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 28, 2013 #2

    PeterO

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    Homework Helper

    I have not checked your calculations closely but a couple of things you may look for.

    Your answer is very close, I wonder if there is:
    (a) some rounding errors?
    (b) the setter used g = 9.8 rather than 9.81?
    (c) the setter used g = 10 rather than 9.81?
     
  4. Jan 28, 2013 #3
    Yeah your right. Thanks buddy!
     
  5. Jan 28, 2013 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You say in your post "I chose positive going down". If you do that, and the ball were thrown up, [itex]v_0[/itex] must be negative.
     
  6. Jan 28, 2013 #5

    PeterO

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    Homework Helper

    I cat and pasted from original

    y=-6m/s*t1 + 4.90m/s^(2) *t1^(2)

    he had -6 ??
     
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