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Throwing a ball upward

  1. Jan 18, 2010 #1
    Hello everyone. I'm sorry for asking such basic questions without offering much in a solution attempt, but the problem is - I have no idea how to answer them! I'm obviously a non-scientist and my math skills aren't so great either. But I would like if someone could simply show me HOW to answer these questions, so I could find the answers myself:

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

    "A ball is thrown straight upward at a velocity of 50 m/s:
    (1) How long will it take for the ball to stop rising?
    (2) How far up has the ball traveled in this time?
    (3) What is the ball's acceleration during this time?"


    2. Relevant equations

    d = 1/2gt2


    3. The attempt at a solution
    I know how to answer #2 if the ball is in freefall, but I don't know what to do if the ball starts at an upward velocity. Ditto for #1 and #3...
     
  2. jcsd
  3. Jan 18, 2010 #2

    rock.freak667

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    Your initial vertical velocity is 50m/s
     
  4. Jan 18, 2010 #3
    Right, that is given in the question. I'm not sure what to do with that...
     
  5. Jan 18, 2010 #4

    rock.freak667

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    In your kinematic equations you have

    v=u+at
    v2=u2+2ad
    d=ut+1/2at2

    u= initial velocity.

    At the highest point, what is the final vertical velocity equal to?
     
  6. Jan 18, 2010 #5
    0 m/s --- So what would I do with that number? (If I should know by now, again I must apologize as I haven't used anything but basic math in years)
     
  7. Jan 18, 2010 #6
    The only two equations you need are
    1/2mv^2 kinetic
    and
    mgh gravitational
    if the ball has reached it's max height then there is no kinetic so if you make these equations equal to each other, the masses will cancel out, you no your speed v, you no gravity g, all you need to account for is height h, which is simple rearranging
     
  8. Jan 18, 2010 #7
    I don't think mass needs to be accounted for, for this question. The question assumes there is no air resistance, and I guess the only relevant factors in figuring out the answers are initial velocity and gravity.

    For the first formula I get: 1/2(50)^2 = 1250 J

    So, what can I do with that number? Should it help me figure out the height?
     
  9. Jan 18, 2010 #8
    Set that formula equal to the gravatational potential equation which is mgh but it's like I said, m is irrelevant and u do no g, all you need is h
     
  10. Jan 18, 2010 #9
    1/2mv^2=mgh
    work from there
     
  11. Jan 18, 2010 #10
    1250=gh
    1250=(9.8)(h)
    1250/9.8=h
    h=127.5m

    Is that correct? If so, how then would I get the time to reach maximum height?
     
  12. Jan 18, 2010 #11
    I don't have a calculator but the equation is right
    height is equal to distance
    use the equation d=v*t since you know 2 of the 3 variables

    t=d/v try it and you should be done
     
  13. Jan 18, 2010 #12
    Be sure to post when your done so I know when to stop lol
     
  14. Jan 18, 2010 #13
    127.5/50=t
    t=2.55

    The last question is "What is the ball's acceleration during this time?" - What should I do to figure that out?

    Also, is there a way to figure out (t) before figuring out (d)? I ask because that question that asks for (t) comes first...
     
  15. Jan 18, 2010 #14
    a=v/t
    now that you have t and have v aswell, just sub in the appropriate values and you are done. I'm done for the day but that should be enough to finish the rest of this question
     
  16. Jan 18, 2010 #15
    If anyone else could offer one last bit of advice:

    Is there a formula that I can use to find (t) before finding (h), if all that is known is the initial velocity?
     
  17. Jan 19, 2010 #16

    rock.freak667

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    The kinematic equations

    v=u+at
    v2=u2+2ad
    d=ut+1/2at2

    u= initial velocity.
    v=final velocity

    At the highest point, the final vertical velocity is zero.
     
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