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Kinematics in Two Dimensions

  1. Mar 22, 2015 #1
    1. The problem statement, all variables and given/known data
    I need to find velocity v from the drawing.

    2. Relevant equations
    Equations of kinematics

    3. The attempt at a solution

    I found that v=12.134 m/s . I need to know is it correct.

    image.jpg
     
  2. jcsd
  3. Mar 22, 2015 #2

    Delta²

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    Your mistake should be in the calculation of s2. Since you use the equation [itex]h=1/2gt_{2}^2[/itex] seems like you ignore the fact that the particle has vertical velocity [itex]v_{y}=gt_{1}/2[/itex] at that point. So it should be [itex]h=v_{y}t_{2}+1/2gt_{2}^2[/itex]
     
  4. Mar 22, 2015 #3
    I get a different answer.

    Maybe you can verbally explain your reasoning?
     
  5. Mar 22, 2015 #4
    Delta means this:in order to find your time for s2, you find the time it takes to fall a distance h under gravity with an initial velocity (in the y direction) of zero. This will give you the wrong time. You need to account for the initial vertical velocity. Think: which takes longer to hit the ground: a ball shot straight forward or a ball shot upward at an initial positive angle above the horizontal?
     
  6. Mar 22, 2015 #5
    I think it's a mistake to combine s1 and s2 the way you are doing because they depend on t^2.

    The standard way to solve a problem like this is to break into two parts: Vx and Vy and use t as the common variable.
     
  7. Mar 23, 2015 #6
    I understand my mistake, it should be h=vyt2+1/2gt22

    Here's what I get:

    image.jpg

    Then I tried to solve it:

    image.jpg

    I think this answer is wrong. Please help!

    By the way, I did solve this equation using Wolfram Alfa and got v=6,88 m/s.
     
    Last edited: Mar 23, 2015
  8. Mar 23, 2015 #7

    Delta²

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    You got drown in a spoon of water :D, you forgot the square root in the last step, the velocity is the square root of 47.4.

    But i am not quite sure the initial equation ([itex]sg=2v_x^2+...[/itex] is correct can u show us how u derive it.

    Ok i think it must be correct cause solving the problem with the "usual" aaproach described briefly by paisiello2 gives the same result.
     
    Last edited: Mar 23, 2015
  9. Mar 23, 2015 #8

    ehild

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    There is a much simpler method based on the trajectory of the projectile. The horizontal displacement (x) is x = Vo cos(α) t , the vertical displacement y - h= Vosin(α) t -g/2 t2. Eliminate the time t, you get the equation y(x) for the whole trajectory:
    ## y = h + \tan(α) x - \frac{g x^2}{2 V_0^2 \cos^2(α)} ##
    You know both the initial point and the final point of the trajectory: if x=0, y=h=0.17 m and x=s=5 m, when y=0. Alpha is given, α=45°. Substitute the known data, and solve the equation for Vo.
     
    Last edited: Mar 23, 2015
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