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Rise of a ball

  1. Oct 5, 2009 #1
    A softball is thrown straight up at 17.5m/s and caught at 3.60s later.How high does the ball rise?

    What equation would I use for this?
    I know that:
    Vi = 17.5m/s
    t = 3.6s
    and gravity is -9.8
    Can anyone help me get started on this problem?
     
  2. jcsd
  3. Oct 5, 2009 #2
    [tex]s(t)=x_{i}+v_{i}t+\frac{1}{2}at^{2}[/tex]

    Also, the total elapsed time is 3.6s. Remember that this is the time it takes for the ball to reach it's highest point and then return.
     
  4. Oct 5, 2009 #3
    what does xi equal?
     
  5. Oct 5, 2009 #4
    s(-9.8) = xi + 17.5 - 4.9t

    [I think]
     
  6. Oct 5, 2009 #5
    [tex]x_{i}[/tex] just represents your initial position. In your case it is zero.

    Use the equation I posted, this calculates displacement. You know that a = -9.8 and your initial velocity is 17.5, lastly you know your initial height is zero. Put that information in the general equation and then calculate the time where the ball would be highest.
     
  7. Oct 5, 2009 #6
    If the round trip takes 3.6 seconds, then at what time will the ball be at it's highest point?
     
  8. Oct 5, 2009 #7
    Here's a hint. When an object is thrown up, it spends half its time going up and the other half coming back down.
     
  9. Oct 5, 2009 #8
    I'm not getting it.
    whats s stand for?
    when I plug it in it is: s(-9.8) = 0 + 17.5 - 4.9(3.6)
     
  10. Oct 5, 2009 #9
    s(-9.8) = 0 + 17.5 - 4.9(3.6)
    s(-9.8) = 0 + 17.5 - 17.64
    s(-9.8) = 17.5 - 17.64
    ???
     
  11. Oct 5, 2009 #10
    It takes the same amount of time to go up as it does to come back down....
     
  12. Oct 5, 2009 #11
    Would it make more sense if I said [tex]
    f(t)=v_{i}t+\frac{1}{2}at^{2}
    [/tex]

    f(t) is just your output; this equation represents the ball's displacement as a function of time. Also keep in mind that the acceleration is multiplied by [tex]t^{2}[/tex] and your initial velocity is multiplied by t.

    Where are you getting s(-9.8) from?

    [tex]
    f(t)=17.5t-4.9t^{2}
    [/tex]

    Thats your equation. Now do you see why the time you need to substitute into that function is not 3.6?
     
  13. Oct 5, 2009 #12
    No, you're trying too hard.
     
  14. Oct 6, 2009 #13
    Still confused?
     
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