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Calculating Initial speed?

  1. Sep 9, 2007 #1
    Calculating Initial speed????

    1. The problem statement, all variables and given/known data

    A ball is thrown straight up at ground level passes a height of 80.6m in 5.1s.
    The acceleration of gravity is 9.8m/s squared. What was its initial speed? answer in units of m/s


    2. Relevant equations

    What equation is the correct one to use to calculate this

    3. The attempt at a solution


    I tried using the formula tup=viy/g

    and i also tried doing it as a table.... where i calculated the velocity at 5.1 secs to be 15.804 m/s... by dividing 80.6 by 5.1 seconds....

    then by adding( since if i was going ahead in seconds instead of back) 9.8 to that speed at 4.1 secs and gettin 25.604 m/s since
    and goin up thru till 0 seconds.... to get velocity of 65.784

    please help me get thru this... this is my last problem for the weekend
     
  2. jcsd
  3. Sep 9, 2007 #2

    learningphysics

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    80.6 is not the maximum height. There's a kinematics equation you can use directly and solve for v1. Think about the equations you have at your disposal.
     
  4. Sep 9, 2007 #3
    i know 80.6 is not the maximum height but im guessin i cannto use the subtracting gravity from speed.... so there is just one kinematics equation that i can use and plug in the data i have that will help me solve this
     
  5. Sep 9, 2007 #4

    learningphysics

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    What is the equation you can use?
     
  6. Sep 9, 2007 #5
    im not sure what equation i can use which i would plug in the 5.1 seconds and the 80.6m as well as the acceleration of gravity to get my initial speed... tehre is one that i can put all 3 into?
     
  7. Sep 9, 2007 #6

    learningphysics

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    Yup there is. what are your displacement formulas?
     
  8. Sep 9, 2007 #7
    what displacement formula lets me take into account gravity
     
  9. Sep 9, 2007 #8
    so would the formula be

    Vf=Vi + at

    where i solve for Vf as 80.6/5.1? and plug the gravity into the equation for the value of A?
     
  10. Sep 9, 2007 #9

    learningphysics

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    No it's a formula that has d... displacement
     
  11. Sep 9, 2007 #10
    would it be..... Vf squared = Vi squared + 2 A D

    where Vf is equal to 80.6/5.1

    A = -9.8m/s

    D = 80.6 m

    ?
     
  12. Sep 9, 2007 #11

    learningphysics

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    Not that equation. The equation has vi, d, t and a. No vf.
     
  13. Sep 9, 2007 #12
    damn i thought that was the it...

    ok so my only other equation that i think is

    D = (Vi t) + 1/2 a t squared

    so D would = 80.6

    a would equal -9.8

    would both the t equal 5.1???

    then solve for Vi to get the initial speed?
     
  14. Sep 9, 2007 #13

    learningphysics

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    exactly.
     
  15. Sep 9, 2007 #14
    ok however... if i solve for that i get the initial velocity to be negative????

    that doesnt make sense
     
  16. Sep 9, 2007 #15
    i get 80.6 = Vi(5.1) + 1/2(-9.8 * 5.1) squared

    then i get

    80.6 = Vi(5.1) + 1/2(-49.980) squared
    80.6=Vi(5.1) + 1/2(2498.00)
    80.6=Vi(5.1) + 1249
    -1168.400 = Vi(5.1)
    -229.098 = Vi

    that does not seem right to me
     
  17. Sep 9, 2007 #16

    learningphysics

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    It should be:
    [tex]\frac{1}{2}gt^2[/tex]

    not
    [tex]\frac{1}{2}(gt)^2[/tex]

    only t is squared. not g.
     
  18. Sep 9, 2007 #17
    or do i only square the time and nto the acceleration of gravity?
     
  19. Sep 9, 2007 #18
    o sorry u posted faster that i typed that question


    so is it (1/2g)(t squared)???
     
  20. Sep 9, 2007 #19

    learningphysics

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    yup.
     
  21. Sep 9, 2007 #20
    so would the initial speed be equal to 40.794m/s in this particular problem
     
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