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Given an initial position and velocity of a receiver, find the velocity of a ball

  1. Oct 28, 2006 #1
    I have a question involving kinematics and physics in a plane. The question is as follows:

    I am working with the two fundamental kinematics equations:

    Code (Text):
    x = v_ix t

    y = v_iy t - 1/2 gt^2
    (An underscore, _, indicates subscripts. A caret, ^, indicates superscripts.)

    Since the final positions of both the receiver and the ball must be identical, I have tried endlessly to modify equations which both involve position (that is, isolating x and then making one formula equal to another). However, some other variable (usually time) always gets in the way of my solving the equation. Any hints which would set me on the right track are appreciated.

    Thanks.
     
  2. jcsd
  3. Oct 28, 2006 #2

    arildno

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    As with most of these problems, the confusion arises because the student does not develop a sufficient notation (or he develops an unclear notation, which is not the case here).

    Relative to Fred's position, Doug has the position as a function of time:
    [tex]x_{D}(t)=6t+20[/tex]
    Again, relative to Fred, the ball has the coordinate positions:
    [tex]x_{B}(t)=v_{x}t=v_{0}\cos(\theta_{0})t, y_{B}(t)=v_{y}t-\frac{gt^{2}}{2}=v_{0}\sin(\theta_{0})t-\frac{gt^{2}}{2}[/tex]
    where [itex]v_{0}[/itex] is the initial speed you are to find, and [itex]\theta_{0}[/itex] is the initial angle (given as 40 degrees).

    Thus, essentially, you are to solve the system of equations for [itex]v_{0}[/itex], the other unknown being the time t:
    [tex]6t+20=v_{0}\sin(\theta_{0})t[/tex]
    [tex]v_{0}\sin(\theta_{0})t-\frac{gt^{2}}{2}=0[/tex]
    knowing that [itex]t\neq{0}, v_{0}>0[/itex]
    Agreed?
     
    Last edited: Oct 28, 2006
  4. Oct 28, 2006 #3
    First of all, thank you arildno for your response.

    Everything you said looks good. Here's what I did:

    Since:

    [tex]6t+20=v_{0}\sin(\theta)t[/tex]

    and

    [tex]v_{0}\sin(\theta)t=\frac{gt^{2}}{2}[/tex]

    I calculated [tex]6t+20=\frac{gt^{2}}{2}[/tex]

    From the quadratic formula, I got roots of -1.50 and 2.72 (which seem reasonable, considering the expected parabola of the projectile).

    Using t = 2.72 s, I plugged the obtained value of t into the equation:

    [tex]y=0=v_{0}\sin(\theta)t-\frac{gt^{2}}{2}[/tex]

    So:

    [tex]v_{0}\sin(\theta)t=\frac{gt^{2}}{2}[/tex]

    [tex]v_{0}\sin(\theta)=\frac{gt}{2}[/tex]

    [tex]v_{0}=\frac{gt}{2\sin(\theta)}[/tex]

    Plugging everything in, I get a value for [itex]v_{0}[/itex] of 20.76.

    When I submit this value, it claims it is incorrect. Does anyone see any errors along the way?
     
  5. Oct 28, 2006 #4

    Office_Shredder

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    Well, it looks like he made a typo:

    [tex]6t+20=v_{0}\sin(\theta_{0})t[/tex] should actually be [tex]6t+20=v_{0}\cos(\theta_{0})t[/tex]
     
  6. Oct 28, 2006 #5
    Ahhh, so it should.

    So now I have:

    [tex]6t+20=v_{0}\cos(\theta)t[/tex]

    and

    [tex]v_{0}\sin(\theta)t=\frac{gt^{2}}{2}[/tex]

    So:

    [tex]v_{0}=\frac{6t+20}{\cos(\theta)t}=\frac{gt}{2\sin(\theta)}[/tex]

    I'll try this and see what happens.

    Thanks, Shredder.
     
  7. Oct 28, 2006 #6
    This time, the quadratic equation yielded roots of -1.41 and 2.43.

    Plugging t = 2.43 s into:

    [tex]y=0=v_{0}\sin(\theta)t-\frac{gt^{2}}{2}[/tex]

    or, as above:

    [tex]v_{0}=\frac{gt}{2\sin(\theta)}[/tex]

    Gives [itex]v_{0}=18.5 m/s[/itex]

    Which is (finally) correct. Thanks to everyone who helped.
     
  8. Oct 29, 2006 #7

    arildno

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    Arrgh, grumph..yet another embarassment. Sorry about the typo in the system of equations at the end of my post..:redface:
     
  9. Oct 29, 2006 #8
    To be honest, I learned more by working through the problem after your suggested solution than I would have learned had the answer been correct.
     
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