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1D Kinematics Question

  1. Jan 11, 2008 #1
    Hi,
    I have a physics problem here that I'm quite stuck on. Maybe I'm making it harder than it seems:

    A ball is thrown upward with an initial speed, Vo. When it reaches the top of its flight, at a height h, a second ball is thrown at the same initial velocity. Find the height where the paths cross.

    No number values are given, so I'm thinking I should be using and manipulating the formula x=xo+vot+0.5at^2 to get a variable answer. I know xo is zero since that's the starting height of both balls and acceleration is gravity (9.81).
    Also, I graphed (displacement vs time) the function assuming that the parabola is even and therefore max height is reached at 0.5 total time. And, the paths crossing would be at 3/4 total time...however, I'm not sure if that's theoretically right.

    Any help would be great! Thanks
     
    Last edited: Jan 11, 2008
  2. jcsd
  3. Jan 11, 2008 #2

    dynamicsolo

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    There is more than one approach for this problem, but this might be the simplest. All you really care about it that the first ball has just come to a stop at a height H, when the second ball is thrown. If you start counting t = 0 from that moment, you have the first ball starting (instantaneously) from rest at x1 = H, and the second ball starting with initial velocity vo from x2 = 0. You will also want to figure out what H is, in terms of vo and g -- how do you do that?

    Write a (one-dimensional) kinematic equation for each ball to express x1 and x2 as functions of time. You want to know at what point they have the same height; setting your equations for x1 and x2 equal will let you solve for the time when that happens. You can then substitute that time into either equation to find the position, measured from where they were thrown, where they pass each other.

    There are perhaps more "elegant" ways to solve this, but this is probably the best while you are getting used to kinematics. (I'll describe one of them later...)
     
    Last edited: Jan 11, 2008
  4. Jan 12, 2008 #3
    Thanks for your response!
    So, I started working on the problem with your advice:

    initially:
    x1=h or x1=h+vot+0.5gt^2 where t=0 and vo=0

    x2=xo or x2=vot+0.5gt^2 so at t=0, x2 would then be 0

    In terms of finding what h is in terms of vo and g (this is where I'm a bit confused)
    could i find this by manipulating my initial formula for x1?
     
    Last edited: Jan 12, 2008
  5. Jan 12, 2008 #4

    dynamicsolo

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    Quick question here: are you calling upward or downward positive? This will affect your acceleration term (although, as it will turn out, not the solution).

    The position equation alone won't help you do that because it doesn't contain information on what the maximum height would be. You can use the velocity equation for a ball being thrown upward to find out when the ball reaches its peak, then use the position equation to find out what h is, in terms of vo and g. (Hint: what is the vertical speed of the ball at the top of the climb?)
     
  6. Jan 12, 2008 #5
    oh gosh, i forgot about the coordinate setting for this (sorry)
    assuming upward is positive, the acceleration would then be negative right?

    when the ball reaches it's peak, I know that velocity is zero
     
  7. Jan 12, 2008 #6

    dynamicsolo

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    Correct.

    What is the velocity equation for a ball being thrown upward? At what time does its velocity reach zero? You can then use your position equation to calculate what h is equal to.
     
  8. Jan 12, 2008 #7
    ok, lets see so far

    v(t)=vo-gt
    when velocity is zero v(t)=0
    t=vo/g

    substituting it into x1=h=vot-0.5gt^2
    i get h to be 0.5(vo^2/g)


    ps: thanks for your help and patience...physics obviously isn't my best subject
     
    Last edited: Jan 12, 2008
  9. Jan 12, 2008 #8

    dynamicsolo

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    All right, so we'll re-write your equation for the first ball using this result:

    x1 = [0.5(vo^2/g)] + 0·t - 0.5g(t^2) ,

    replacing h and using vo = 0 (and fixing the sign on the acceleration term) for this ball which is about to start falling from its maximum height. (Again here, we are calling t = 0 the moment when this ball starts to fall and the second ball is just being tossed upward.) The position of the second ball, as you've said, is

    x2 = 0 + vo·t - 0.5g(t^2) .

    So we want to set these positions equal to find the moment when the two balls are at the same height. You can then use this time to find the height at which this occurs.

    I don't see why you say that -- you're doin' good so far...
     
  10. Jan 12, 2008 #9
    all right! if i did the math right (hopefully) to wrap up this problem

    I've come to the answer of 3vo^2/8g
     
  11. Jan 12, 2008 #10

    dynamicsolo

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    Very good! (And you said this wasn't your best subject...) It also turns out that your statement (calculation? physical intuition?) that this occurs 3/4 of the way through the flight of the first ball is correct; this only is true because the motion of both balls is identical except for the offset in "launching time".

    I mentioned there are other approaches to this problem. One is to picture yourself falling alongside the first ball and watching what both balls are doing, in what is called an "accelerating (or non-inertial) reference frame". Since both balls are in the same gravitational field, both are accelerating downward at g, and so are you. You see the first ball hanging stationary alongside you, while the second ball is coming up at you with a constant speed vo. Because the two balls started with a separation of

    h = 0.5(vo^2/g) ,

    the time it takes for that second ball to rise to meet you is

    t = h/vo = 0.5(vo/g) .

    (Note that this is half the time required for either ball to reach the maximum height and so indeed occurs 3/4 through the first ball's travels.) While this is happening, you also see the ground accelerating upward toward you at g. In the time that has passed, you and the first ball will have drawn closer to the ground by

    d = 0.5 · g · (t^2) = 0.5 · g · (0.5^2) · (vo/g)^2
    = 0.125 · (vo^2/g) ,

    now placing you and both balls at a height

    h - d = (3/8) · (vo^2/g) .

    You may encounter accelerating reference frames later in your course. With experience, they can be quite handy for solving some kinds of problems (or checking solutions) .
     
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