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2D Kinematics Help

  1. Jan 31, 2008 #1
    [SOLVED] 2D Kinematics Help

    Basically I want to make sure that I have worked this correctly and find out what I did wrong in Part 3, as my solution is not logical.

    Problem Consider a ball thrown up from the ground. It passes a window in the time interval .251s. The distance across the window is 1.49m.

    1) Find the average speed as the ball passes the window.
    2) What is the magnitude of the decrease of the velocity across the window?
    3) If the ball continues its path upward without obstruction, find the travel time between the top of the window and the ball's maximum height.

    My Work

    1)V = (Vo + Vf)/2 = (Xf + Xi)/t (This correspondence was the only thing I could come up with to solve this problem)

    5.936 = (Vo + Vf)/2
    2(5.936) - Vf = Vo

    Vf = Vo +at
    Vf = 2(5.936) - Vf + at
    2Vf = 2(5.936) + at
    Vf = 5.936 + (at/2)
    Vf = 5.936 + (-9.8t/2)
    Vf = 5.936 + ((-9.8 * .251)/2)
    Vf = 4.706 m/s

    2)Magnitude = absolute value of Vf - Vo

    Vo = 2(5.936) - 4.706
    Vo = 7.166

    |4.706 - 7.166| = 2.46 m/s

    3) x = (Vf^2 - Vo^2)/2a

    x = (0 - 4.706^2)/(2 * -9.8)
    x = 1.12992

    dx = Vo*t + .5 at^2
    dx = 4.706t + .5 * -9.8t^2
    1.12992 = 4.706t - 4.9t^2
    1.12992 = t(4.706 - 4.9t)
    1.12992/t = 4.706 - 4.9t
    (1.12992/t) + 4.9t = 4.706
    1.12992 + 4.9t = 4.706t
    1.12992 = -.194t
    t = -5.82455

    Obviously time shouldn't be negative. Did I do parts 1 & 2 correctly? Also, where did I go wrong with part 3?

    Thanks in advance for any help.
  2. jcsd
  3. Jan 31, 2008 #2
    I think you better double check the equations in bold. You're going to be solving a quadratic, and thus you should have two times, one negative and one positive.
  4. Jan 31, 2008 #3
    I'm being very dense for some reason. I have no idea what is wrong with it. :)
  5. Jan 31, 2008 #4
    How did you get rid of the "t" under "1.12992" ? If you multiplied both sides by "t", then how come the 4.9 didn't get any of that action?
  6. Jan 31, 2008 #5
    Wow. I can't believe I did that. I'll just blame the internet :D

    Is there a better equation to use for "t"?

    Also, do the previous two look okay?
  7. Jan 31, 2008 #6
    I don't believe there's a better equation, you'll just have to solve for the quadratic. I'll look over the other ones, but it may take me a minute.
  8. Jan 31, 2008 #7
    1) look again at the question. (vf+vo)/2 *is* the average speed across the window
    2) you had the right form of the answer while you were trying to work out 1)
    3) v=0 at max height - work from the top of the window to get there.


    Last edited: Jan 31, 2008
  9. Jan 31, 2008 #8
    The average speed is given by [tex] v_{AVE} = \frac{distance}{time} [/tex]

    Simply plug in the values.

    Your answer for number 2 is correct.

    Let me know what you get for number 3.
  10. Jan 31, 2008 #9
    For #3 I think you can use the equation:

    [tex] v_2 = v_1 + at[/tex]


    [tex]v_2 = 0 m/s [/tex] (at max height velocity is 0 as antenna guy notes)
    [tex] v_1 = 4.706 m/s [/tex] (I'm assuming the calculation for this is correct)
    [tex] a = -9.8 m/s^2[/tex]

    Then just solve for 't'.
  11. Feb 1, 2008 #10
    "1) look again at the question. (vf+vo)/2 *is* the average speed across the window"

    The problem is that I'm not given V or Vf, so I couldn't use it to solve the problem. I only have Vo because I assume that it starts at rest.

    After solving the quadratic, I got X1 = .48, X2 = .4804. I then threw it into a Quadratic Equation Solver and it agreed with that outcome.

    The equation given by chocokat (which I didn't see until just now LOL) gives t = .4802. So should I be safe with .48s?

    As for this equation:

    "The average speed is given by Vave = d/t", would this work when the object has an unknown speed before it enters the field of view?

    I'm sorry I'm being such a pain. I just have almost no confidence in my answers.
  12. Feb 1, 2008 #11
    Your assumption is incorrect, but your answer to (vf+vo)/2 is. v=dx/dt, and the window could be any height above the ground. vis:


    BTW - dx=xf - xi (not " +")

    The amount of work you put into answering 2) is commendable (although you did most of the work while trying to solve 1)). In the future, you might consider that:


    I'm not sure you realize what you did to get your answer to part 3, so I'll try to describe what I see:
    By substituting in 0 for Vf, I assumed you changed what Vf and Vo stood for. Vf is now v at max height, and you solved for the height above the window.

    The quadratic approach you eventually used to resolve t from that point looks reasonable, but there is an easier way:

    Let v_f be the velocity at the top of the window, and v_m be the velocity at max height.







    Yes, it does work - and you can figure out the speed as the ball enters the window.




    Keep chugging - try to understand what the equations mean, and the confidence will come.


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