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Projectile Motion: How tall is the building?

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball thrown horizontally at 12.4m/s from the roof of a building lands 20.2m from the base of the building. How tall is the building?

    2. Relevant equations
    I think I'm supposed to use these:
    horizontal motion: x=x0+vx0t
    vertical motion: y=y0=+vy0(t)-.5at^2

    3. The attempt at a solution
    So this is what I think goes where:
    x0= 0
    y0= 0
    a= -9.8
    I plugged them in but I'm not sure if they're even in the right spot. After I plug them in I have no idea where to go from there. Please help.
  2. jcsd
  3. Feb 3, 2015 #2
    Horizontal motion is not accelerated or decelerated here. So, if you know the distance horizontally and the velocity horizontally, you can calculate time. From there you can figure out the height

    In your solution, you put vy0 as the 12 m/s but it should be vx0
  4. Feb 3, 2015 #3
    I got the time as approx. 1.6290. Is that right? I plugged the time into the vertical equation and got approx. y=-5.8067
  5. Feb 3, 2015 #4


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    Right time. How did you manage to get that y value though? Show what you plugged into what.
  6. Feb 3, 2015 #5
    I have y=y0+vyo(t)+.5at^2
    Then I plugged in y=0+12.4t+.5(-9.8)t^2
    Which then becomes y=12.4t+.5(-9.8)t^2
    I then plugged in the time, so y=12.4+.5(-9.8)(1.6290)
  7. Feb 3, 2015 #6


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    12.4 m/s is vx0. What's vy0? It says 'thrown horizontally'.
  8. Feb 3, 2015 #7
    So it's 0 m/s instead of 12.4?
  9. Feb 3, 2015 #8
    For vy0, yes it is 0 m/s.
  10. Feb 3, 2015 #9
    So it's 0 m/s instead of 12.4 m/s?
  11. Feb 3, 2015 #10
    Sorry I didn't mean to post that twice. So I got approx. 13.0034 for the height of the building. Is that correct?
  12. Feb 3, 2015 #11
    Looks good ;)
  13. Feb 3, 2015 #12
    Thank you guys so much!
  14. Feb 3, 2015 #13
    NP. Glad you got it :P
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