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Acceleration of free falling object on another planet

  1. Sep 26, 2007 #1

    ~christina~

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    [SOLVED] Acceleration of free falling object on another planet

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

    A ball is shot vertically upward from the surface of another planet.
    A plot of y versus t for the ball is shown in the figure below, where y is the height of the ball above its starting point and t = 0 at the instance the ball is shot.

    The figure’s vertical scale is set by ys = 30.0 m.

    (a) Write expressions for the displacement and velocity of the ball as functions of time.

    (b) What is the magnitude of the free-fall acceleration on the planet?

    (c) What is the magnitude of the initial velocity of the ball?

    ~I've attatched a pic of the graph.

    2. Relevant equations

    I have to get the acceleration.


    3. The attempt at a solution

    a.) I need help on how to derive a equation from that graph and I don't know what the ys= 30m scale is that they describe.
    I do know that since it is x vs. t graph I can get the v from that.


    b.) would I solve the equation I derive or use another equation for getting the acceration due to gravity for this? I don't know if I get that from a equation that I derive.


    ~Thanks~:confused:
     

    Attached Files:

  2. jcsd
  3. Sep 27, 2007 #2

    Kurdt

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    For part a all you will need are the standard kinematic equations. Then for part b you would solve for acceleration using some selected values from the graph. You will have to substitute for the initial speed in terms of acceleration and then once you've found the acceleration you can find the initial speed.

    Just because it says its on a different planet does not mean that you need new equations. The only difference is the magnitude of the acceleration at the surface.
     
  4. Sep 27, 2007 #3

    ~christina~

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    well for part a.) would it be using these equations?

    vxf = vxi + ax*t --for velocity as a function of time
    and

    xf= xi + vxi*t + 1/2 ax*t^2---position as a function of time

    That seems deceptively simple if it's right since I didn't do anything to the equations but do I need to do anything to them??
    Do I need to combine them??
    ___________________________________________________
    for part b.) I'm not quite sure what you mean to "substitue for the initial speed in terms of acceleration."

    Do you mean that I get velocity as the slope of the graph for d vs t
    then use that and plug into which equation and for what ?

    ex. If I choose d= 15 and d= 10
    and t= 0.6 t= 1

    and I find the velocity which would be the inst velocity right?

    so...
    v= (15-10)/(1-.6)
    that would = 12.5m/s

    but wouldn't that be the average velocity??

    I don't know what to do with that number if I did that correctly.


    Help ~Thank You ~
     
  5. Sep 27, 2007 #4

    Kurdt

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    For part b I meant you have two equations with two unknowns. You will rewrite the velocity equation in terms of initial velocity and substitute into the distance equation and then solve for acceleration. Now to get rid of the final speed term you will have to choose a point on the graph where the final speed is zero. There is only one place where that happens. Once you have the acceleration you can then go back and find the initial speed.

    I hope that is clear.

    [tex] v= u +at [/tex]

    [tex] x = ut+\frac{1}{2} at^2 [/tex]

    Hint: [tex] u = v - at [/tex]
     
  6. Sep 28, 2007 #5

    ~christina~

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    see below
     
    Last edited: Sep 28, 2007
  7. Sep 28, 2007 #6

    ~christina~

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    Well doing that I get...

    vxf= vxi + ax*t= vxi= vxf-ax*t

    xf= xi + vxi(t) + 1/2(ax)t^2

    plugging in the vxi equation into the xf equation gives me....

    xf= xi + vxi(t) + 1/2(ax) t^2

    xi= 0

    xf= (vxf- ax*t)t + 1/2(ax)t^2

    xf= vxf*t - ax*t^2 + 1/2(ax)t^2

    ~now I have a question....you said to solve for ax and then find a point where final speed = 0

    first wouldn't that be at d= 25 and t= 2.5?
    I know that it should be 0 since the graph peaks at that point but I don't get zero when I divide the numbers (v= d/t)...or is it at the end of the graph since it's the "final velocity"
    so my second guess and a more likely one in my opinion is that the final v= 0 is at

    d= 0 while t= 5s Is this okay?

    secondly how would I solve for ax if ax would cancel out since you multiply it by t so you get
    xf= vxf*t - ax*t^2 + 1/2(ax)t^2 = vxf *t -.5 If I'm not wrong...

    ~Thanks~
     
  8. Sep 28, 2007 #7

    Kurdt

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    Hey christina,

    You're ok up until the point where you are choosing the final velocity to be zero. remember the velocity on a distance time graph is the gradient of the curve. So velocity is zero when there is a stationary point of the curve. That means your first guess was correct. Final velocity is zero when vertical distance is 25m and time is 2.5 s. The reason why you don't get zero when you divide the distance by the time is because you are calculating average speed.

    Secondly, [itex]\frac{1}{2}at^2 - at^2 = -\frac{1}{2}at^2[/itex]

    Ok?
     
  9. Sep 28, 2007 #8

    ~christina~

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    Continuing from my previous post...

    vxf= vxi + ax*t= vxi= vxf-ax*t

    xf= xi + vxi(t) + 1/2(ax)t^2

    plugging in the vxi equation into the xf equation gives me....

    xf= xi + vxi(t) + 1/2(ax) t^2

    xi= 0

    xf= (vxf- ax*t)t + 1/2(ax)t^2

    xf= vxf*t - ax*t^2 + 1/2(ax)t^2

    xf= vxf*t - 1/2(ax)t^2

    xf= 25m
    vxf= 0

    25m= -1/2(ax)t^2

    ax= -8m/s^2 ===> this would be the magnitiude for the freefall acceleration on the planet.

    now to get the initial velocity I plug that ax value into

    vxf = vxi + ax*t

    or

    xf= xi + vxi*t + 1/2 ax*t^2 ?

    I'm not sure b/c I'm confused as to what t value to use and what xf value to use as well as the xi ...

    my thoughts are that
    xf= 25m
    t= 2.5s
    ax= -8m/s^2
    xi= 0

    so plugging in to the 2nd equation to find vxi since I'm not sure about the first equation and what t would I use there...

    xf= xi + vxi*t + 1/2 ax*t^2
    25m= 0 + vxi*(2.5s) + (1/2*(8m/s)*2.5s^2)
    1m/2.5s= vxi

    vxi= 0.4m/s


    Is this alright??

    Thanks Kurdt
     
  10. Sep 28, 2007 #9

    Kurdt

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    Personally I'd go with vxf = vxi + ax*t

    You can use the same values you used to find the acceleration. So vxf will be zero and t will be 2.5s.

    The method you have used is fine except you have made a slight error. Remember that the acceleration is -8ms-2. So you would end up with [itex]\frac{50}{2.5} = a[/itex].

    Other than that its fine. :smile:
     
  11. Sep 28, 2007 #10

    ~christina~

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    oh..but the reason I used 8.0 instead of -8.0 was that I think that wouldn't the velocity have to be negative if the acceleration is negative but I guess that only applies if I'm tossing something


    Well I got vxi= 20m/s




    THANKS alot for your help Kurdt !! =D
     
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