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Help With finding initial vertical speed

  1. Sep 27, 2013 #1
    1. The problem statement, all variables and given/known data

    A Ball is thrown upwards. What is its initial vertical speed? The acceleration of gravity is 9.8 m/s2 and maximum height is 7.2 m ,. Neglect air resistance.


    2. Relevant equations

    Vfinal squared = velocity initial squared + 2g times change in y

    3. The attempt at a solution

    velocity final squared = velocity initial squared + 2 (-9.8)(7.2)
    VF^2= VI^2 + -141.12
    what now?/
     
  2. jcsd
  3. Sep 27, 2013 #2

    Doc Al

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    What's the final velocity? (When it reaches the max height.)
     
  4. Sep 27, 2013 #3
    the final velocity is not given. all that is given is v0, max height is 7.2 meters, and that the acceleration of gravity is 9.8 m/s/s/.
     
  5. Sep 27, 2013 #4
    is the finaly velocity 0?
     
  6. Sep 27, 2013 #5

    Doc Al

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    Of course. :wink:
     
  7. Sep 27, 2013 #6
    ohh ok... so would i use the same formula i used above to find the initial vertical speed?
     
  8. Sep 27, 2013 #7
    what equation should i use to calculate the initial vertical speed?
     
  9. Sep 27, 2013 #8

    Doc Al

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    Just use that equation you started with and solve for the initial speed.
     
  10. Sep 27, 2013 #9
    how do i set it to be that velocity initial squared is on the left side of the equal sign, and velocity final squared is on on the right side?
     
  11. Sep 27, 2013 #10

    Doc Al

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    The final velocity is zero, so your equation becomes:

    0 = Vi2 - 141.12

    So how to get Vi2 alone? How can you get rid of that - 141.12?
     
  12. Sep 27, 2013 #11
    you add 141.12 to the 0 and then find square root of 141.12?
     
  13. Sep 27, 2013 #12

    Doc Al

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    Sure! Basic algebra.
     
  14. Sep 27, 2013 #13
    so the initial vertical speed would be 11.8793939239 m/s ?
     
  15. Sep 27, 2013 #14

    Doc Al

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    Yes. But please round off to a reasonable number of significant figures. (2 or 3 is plenty.)
     
  16. Sep 27, 2013 #15
    ok, i have one more question i need help with....
     
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