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SUVAT motion question - Ball thrown vertically

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    A cricket ball is thrown vertically upwards and returns in 4.2s. Find (a)The initial velocity of the ball

    (b)The maximum height reached


    2. Relevant equations
    v=u+at
    v^2=u^2+2as
    s=(u+v)/2 multiplied by t
    s=ut+1/2at^2


    3. The attempt at a solution

    s=?
    u=?
    v=?
    a=9.81ms^-2 (negative travelling upwards)
    t=4.2s

    I could not start the question because to use SUVAT equations at least 3 of the 5 letters must have values to be put into a equation.
     
  2. jcsd
  3. Sep 28, 2011 #2
    Start with

    v = v_0 + at
    Where v = end velocity, v_0 initial velocity, a = g = gravity and t = time

    If you consider v to be the velocity at the top of the throw then v = 0.
    Thus 0 = v_0 + at
    v_0 = -at, You know the acceleration and t = half the total time. So there you get the initial velocity and rest becomes simplier.
     
  4. Sep 28, 2011 #3

    Doc Al

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    Staff: Mentor

    There are two equations you can use.
    First fill in these:
    s = ?
    v = ?
    (You have enough information to fill in these values.)
     
  5. Sep 28, 2011 #4
    THINK!!!
    What is the velocity at the max height?
     
  6. Sep 28, 2011 #5
    s= I'm really not sure how to found out the displacement
    v=0 [ as it has stopped at its maximum point in the air?]
     
  7. Sep 28, 2011 #6
    0 velocity?
     
  8. Sep 28, 2011 #7
    0 = v_0 + 9.81x4.2 ? I am not sure I have done this correctly

    so v_0 = -41.202 <---- I've definitely gone wrong here, the initial speed surely cannot be negative
     
  9. Sep 28, 2011 #8

    Doc Al

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    It looks like you are trying to use this equation: s=ut+1/2at^2

    But you left out a few things. Try again. (But yes, s = 0 for the round trip.)
     
  10. Sep 28, 2011 #9
    Ah, thankyou s now makes sense
     
  11. Sep 28, 2011 #10
    0=u x 4.2 + 0.5 x 9.81 x 4.2^2

    gives me 0 = u x 90.7242

    so u = 0

    I'm really struggling with this question, what error have I made this time?
     
  12. Sep 28, 2011 #11

    Doc Al

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    OK, except that you have the wrong sign for the acceleration. Fix that first.

    No it doesn't. Try again.
     
  13. Sep 28, 2011 #12
    -9.81 as its against gravity, rookie error

    I now got -82.3242 as my answer but I'm not sure how this can be right as it is negative
     
  14. Sep 28, 2011 #13

    Doc Al

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    Good.
    Show how you got that, step by step.
     
  15. Sep 28, 2011 #14
    0=u x 4.2 + 0.5 x -9.81 x 4.2^2

    so 0 = u x -82.3242
     
  16. Sep 28, 2011 #15

    Doc Al

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    OK.
    Not OK.

    Try this:
    0=u x 4.2 + 0.5 x -9.81 x 4.2^2
    which I'll rewrite as:
    0 = 4.2u -(0.5)(9.81)(4.2^2)
    4.2u = (0.5)(9.81)(4.2^2)

    And so on...
     
  17. Sep 28, 2011 #16
    Just confused on where the - sign has come from in the line: 0 = 4.2u -(0.5)(9.81)(4.2^2)
     
  18. Sep 28, 2011 #17

    Doc Al

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    Starting with:
    0 = 4.2u +(0.5)(-9.81)(4.2^2)

    You can move that minus sign from in front of the 9.81 and place it here:
    0 = 4.2u -(0.5)(9.81)(4.2^2)

    (1)(2)(-3) is the same as -(1)(2)(3).
     
  19. Sep 28, 2011 #18
    4.2u = 86.5242

    /4.2 = 20.601 ms^-1 Final answer for (a) I think?
     
  20. Sep 28, 2011 #19
    Right OK, I see how you have done that. Is my answer of 20.601 correct?
     
  21. Sep 28, 2011 #20

    Doc Al

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    Yes. (I'd round it off to no more than 3 sig figures.)
     
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