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Homework Help: How far beyond the fence does the ball strike the ground?

  1. Jun 15, 2008 #1
    The problem statement, all variables and given/known data
    A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.6 s after being hit. Then 3.1 s after reaching its maximum height, the ball barely clears a fence that is 87.5 m from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?

    The attempt at a solution
    (a) 63.5 m

    (b) 16.4 m

    (c) ??

    I got these answers by using H=.5gt^2

    (a) H=.5(9.8)(3.6)^2 = 63.5 m <--max height

    (b) H=.5(9.8)(3.1)^2 = 47.1 m
    63.5-47.1 = 16.4 m <--height of fence

    How do I go about finding part (c) How far beyond the fence does the ball strike the ground?
  2. jcsd
  3. Jun 15, 2008 #2
    Well you can calculate the time it takes to strike the ground after clearing the fence.
  4. Jun 15, 2008 #3
    For part c, you need to calculate horizontal distance, and for this you need to use the fact that the fence is at a distance of 87.5m from hit. Can you calculate the horizontal component of velocity?
  5. Jun 15, 2008 #4
    Would horizontal component for velocity be time? If so would it be 6.7?
  6. Jun 15, 2008 #5
    The horizontal component doesn't change it's constant. So if you have the time it takes to get to the fence and the distance to the fence you can calculate the horizontal component
  7. Jun 15, 2008 #6
    When you say horizontal component what do you mean?
  8. Jun 15, 2008 #7
    I mean the velocity in the horizontal direction. The baseball has a vertical and horizontal velocity. The vertical velocity is changing the horizontal is not changing. Understand?
  9. Jun 15, 2008 #8
    so like V= 87.5/6.7 = 13.1 m/s
  10. Jun 15, 2008 #9
    Yes. But you must now find the distance traveled beyond the fence
  11. Jun 15, 2008 #10
    Ok so would I set up an equation like this x= vt and if thats true how would I know what t equals?
  12. Jun 15, 2008 #11
    Well you know the time it takes to reach the maximum height and the time it takes to reach the fence. So what would you do to find the time to reach the ground beyond the fence?

    Hint: what is the total time?
  13. Jun 15, 2008 #12
    3.6s + 3.1s = 6.7 s to reach the fence then I'm lost. How would I calculate the time to reach the ground? I know that the velocity when it hits the ground is 0 correct?
  14. Jun 15, 2008 #13
    Okay think of it another way. When you throw a baseball up in the air it takes 3 seconds to reach the maximum height then it start comming back down to you. How long would that take?
  15. Jun 15, 2008 #14
    3 seconds?
  16. Jun 15, 2008 #15
    Exactly. Can you apply this to your current problem?
  17. Jun 15, 2008 #16
    use the equation v=xt and solve for x when v is 13.1 m/s and t is 0.5s?
  18. Jun 15, 2008 #17
    You simply have to find the distance it takes to reach max height and double it.
  19. Jun 15, 2008 #18
    You mean the time.
  20. Jun 15, 2008 #19
    V doesn't equal to xt. V=x/t
  21. Jun 15, 2008 #20
    i meant x=vt

    and solve for x like this

    x=94.32 m

    94.32-87.5= 6.82m

    is that right?
  22. Jun 15, 2008 #21
    Yea I figured but I just wanted to make sure
  23. Jun 15, 2008 #22
    No, distance covered to reach the highest point is same as distance covered when coming down from the highest point, since horizontal velocity is constant. Time taken also are the same.
  24. Jun 15, 2008 #23
    I was correcting your english. The distance it takes to reach the maximum height. See what I mean.
  25. Jun 15, 2008 #24
    Yes that is correct. I wouldn't have rounded the velocity up to 13.1. Instead I would have left it till the end of the problem and had my final answer to the correct sig figs because my answer is 6.53 when you don't round.
    Last edited: Jun 15, 2008
  26. Jun 15, 2008 #25
    Oh sorry, bad construction. I meant horizontal distance it takes to reach maximum height of course. Thanks ace.
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