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Throwing a ball: Motion in one direction

  1. Feb 3, 2012 #1
    ~ Throwing a ball: Motion in one direction ~

    1. The problem statement, all variables and given/known data
    The boy throws the ball vertically from the top of the building. The ball travels upward for 4 seconds. a) What was the original velocity of the ball when it was thrown? b) How high above the building does the ball go? When the ball begins its descent, it passes its original position at the top of the building. c) What is the velocity of the ball at this point? d) If the building is 40m, compute the elapsed time of flight from when the ball left the boy's hand. e) At what velocity does the ball hit the ground? f) What is the acceleration of the ball at the peak of its flight? g) What is the acceleration of the ball just before it hits the ground?



    2. Relevant equations
    v = v-original + at
    Delta(x) = V-original(time) + (1/2)at^2
    Ymax = V-original(time) + 1/2 at^2



    3. The attempt at a solution

    a) Used velocity equation instead.
    0 = Vo + (-9.8)(4)
    0 = Vo + (-39.2)
    39.2m/s = Vo

    b) This calculation makes no sense to me. EDIT: Maybe this is better. I still don't see how a boy could possibly throw a ball twice the height of a building he's standing on top of.
    Ymax = Vo(time) + (1/2) at^2
    Ymax = 39.2(4) + (1/2)(-9.8)(16)
    Ymax = 156-78.4
    Ymax = 78.4

    c)
    v = Vo + at
    v = 39.2 + (-9.8)(8)
    v = -39.2 m/s

    d)
    -40 = 39.2t + (1/2)(-9.8)(t)^2
    -40 = 39.2t + (-4.9)(t)^2
    -4.9t^2 + 39.2t + 40 = 0
    4.9t^2 - 39.2t - 40 = 0

    = 39.2 plus/minus sqrt (1536.64 + 784)/9.8
    = 39.2 plus/minus sqrt (2320.64)/9.8
    = 39.2 plus/minus 48.17/9.8
    t = 8.92 seconds

    WHEW.

    e) v = Vo + at
    v = 39.2 + (-9.8)(8.92)
    v = 39.2 + (-87.42)
    v = -48.22
    v = 48.22 m/s

    f) -9.8 m/s^2

    g) -9.8 m/s^2
     
    Last edited: Feb 3, 2012
  2. jcsd
  3. Feb 3, 2012 #2

    Doc Al

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    Re: ~ Throwing a ball: Motion in one direction ~

    It says the ball travels upward for 4 seconds, not that the round trip back to the start took 4 seconds.

    Redo this, as it will affect the other answers.
     
  4. Feb 3, 2012 #3
    Re: ~ Throwing a ball: Motion in one direction ~

    I'm still not sure how this affects the question. Can I assume that since it takes 4 seconds to reach the peak, it will take 8 seconds to fall back down to its original position?

    Physics noob, here.
     
  5. Feb 3, 2012 #4

    Doc Al

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    Re: ~ Throwing a ball: Motion in one direction ~

    Sure. That's one way to do it. Another way is to use another kinematic equation that doesn't involve distance.

    (Do it both ways!)
     
  6. Feb 3, 2012 #5
    Re: ~ Throwing a ball: Motion in one direction ~

    One more question, and thank you for the help by the way.

    How am I to find the displacement, or distance, if I don't have an initial velocity? Using the equation Delta(x) = V-original(time) + (1/2) at^2, I have all of the variables besides Vo and delta(x) - and I assume delta(x) is stated implicitly somewhere in the problem and I'm just missing it -
     
  7. Feb 3, 2012 #6
    Re: ~ Throwing a ball: Motion in one direction ~

    you don't have an initial velocity, but you do have velocity after 4 seconds. When the ball hits the peak, its velocity is equal to zero.
     
  8. Feb 3, 2012 #7
    Re: ~ Throwing a ball: Motion in one direction ~

    Thank you sir, it's much appreciated.
     
  9. Feb 3, 2012 #8

    Doc Al

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    Re: ~ Throwing a ball: Motion in one direction ~

    You first solve for the initial velocity, of course. That's why it's part a!

    If you use the displacement formula to solve for the initial velocity, you must choose an interval for which you already know the displacement. Such as a round trip. But you can also use the velocity formula to solve for the initial velocity.
     
  10. Feb 3, 2012 #9
    Re: ~ Throwing a ball: Motion in one direction ~

    Reworked, but I'm not sure if it's correct. I wish this thing came with an answer key.

    EDIT2: Worked out part c.
     
    Last edited: Feb 3, 2012
  11. Feb 3, 2012 #10

    Doc Al

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    Re: ~ Throwing a ball: Motion in one direction ~

    Looks good now.

    Sure. (But if you catch the trick, this shouldn't require any calculation.)
    Use the same formula. What's the displacement from the initial position?
     
  12. Feb 3, 2012 #11
    Re: ~ Throwing a ball: Motion in one direction ~

    Woot.

    I'm assuming the trick is some sort of law that states if there's no other resistance, an object that travels x meters upward at a velocity will have the same velocity as the initial if it falls the same x meters?

    39.2? Which formula would I be plugging this into?
     
  13. Feb 3, 2012 #12

    Doc Al

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    Re: ~ Throwing a ball: Motion in one direction ~

    Yep. The same speed, not velocity, since the direction is opposite. (Thus the minus sign.)
     
  14. Feb 3, 2012 #13

    Doc Al

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    Re: ~ Throwing a ball: Motion in one direction ~

    That depends on how fast the boy can throw and how high the building is, doesn't it? Of course, this kid is obviously a major league fastball pitcher! (Not very realistic.)

    :yuck:
     
  15. Feb 3, 2012 #14
    Re: ~ Throwing a ball: Motion in one direction ~

    Lol. *facepalm*

    Was I wrong? I'm still trying to figure out which equation I'd use to solve part d. Any hints? I suppose the displacement at the peak is 39.2, but I don't know how that helps me.
     
  16. Feb 3, 2012 #15
    Re: ~ Throwing a ball: Motion in one direction ~

    an important thing about the equations used here is that they only work when acceleration remains constant. Also, acceleration is caused by forces, which means basically that acceleration only happens when something pushes on something else. In this situation, after the ball leaves the kid's hand, there is nothing pushing on the ball except for gravity (if we ignore air resistance). This is also why they make the important distinction of "just before" in g.
     
  17. Feb 3, 2012 #16
    Re: ~ Throwing a ball: Motion in one direction ~

    Another thing I was trying to understand; I know that when objects fall in a non-vacuum, their acceleration is not constant. AKA, after one second, a ball would NOT fall 9.8 meters, but something like 4.

    Arg.
     
  18. Feb 3, 2012 #17

    Doc Al

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    Re: ~ Throwing a ball: Motion in one direction ~

    About the acceleration being zero? Yep, way wrong. What's the acceleration of a projectile at any point in its motion?
    You'd use the same equation as before. (The second in your list.) Displacement (Δx) means: Final position - initial position.
     
  19. Feb 3, 2012 #18
    Re: ~ Throwing a ball: Motion in one direction ~

    that's because air is in the way, basically. It's the same idea as why it's hard to move through water. The water is "pushing" back at you in the form of resistance. Air "pushes" on falling objects in the same way.

    in simple problems like yours, we just forget about it because it makes things a lot more complicated
     
  20. Feb 3, 2012 #19

    Doc Al

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    Re: ~ Throwing a ball: Motion in one direction ~

    First solve the 'simple' problems that ignore complications such as air resistance--you can worry about the harder and more realistic problems later.
     
  21. Feb 3, 2012 #20
    Re: ~ Throwing a ball: Motion in one direction ~

    I don't know. ._. I know the acceleration in the x direction is always constant, I don't know about the y direction.

    Will try.
     
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