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2D Motion part deux

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data

    An object is shot from the origin with a velocity of 40.0 m/s at an angle of 55.0 degrees above the horizontal. What is the location of the object 3.00 seconds later?


    2. Relevant equations

    All 2D motion equations - too many to list Ex:

    [tex] V_x = v_0x + a_x t [/tex]

    [tex] V_y = v_0y + a_y t [/tex]


    3. The attempt at a solution
    What I am having trouble with is the 3.0 seconds later part. I am not sure how to approach this one. I have to find the x displacement and then the angle [tex] \theta [/tex] which is no problem to do when I know [tex] V_y and V_x [/tex].

    So I tried this:

    [tex] \Delta x = 40 cos(55)3.0 = 68.8m [/tex] , but none of the answer choices match this. So I must be doing something wrong. I also can't get a correct answer for [tex] V_y and V_x [/tex] with these I could plug them into the inverse tangent equation and get an angle. To find [tex] V_x [/tex] I do this:

    [tex] V_x = 40 cos (55) = 22.9 [/tex]

    for [tex] V_y [/tex] I do this :

    [tex] V_y = 40 sin (55) - (9.80)(3.0^2) = -55.4 [/tex]

    I am not sure that I am using the time variable correctly.
     
  2. jcsd
  3. Sep 20, 2009 #2

    sylas

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    Science Advisor

    You can solve the horizontal and vertical parts independently.

    Your Δx of 68.8 looks fine.

    Your final equation for Vy is wrong. Are you calculating a velocity, or a distance?

    Under relevant equations, you've given a correct formula for getting a velocity, given initial velocity and acceleration.... but it doesn't have a t2 factor anywhere. The question, of course, requires a distance.

    Do you know an equation for distance, given an initial velocity, an acceleration, and a time?

    Cheers -- sylas
     
  4. Sep 20, 2009 #3
    well there is this one:

    [tex] \Delta x = V_0_x t + (1/2) (a_x) t^2 [/tex]

    but [tex] a [/tex] in the horizontal direction = 0 in this case. So I am back to where I started.

    I am trying to find, where the object is going to end up after 3.00 seconds.

    I need the distance traveled in 3 seconds. And then I need to find the angle that vector makes with the x-axis using the inverse tangent equation. But to use the tangent equation I need to find [tex] V_y and V_x [/tex]

    These are the equations from my book:

    [tex] V_y = V_0 sin \theta - gt [/tex]
    [tex] V_x = V_0 cos \theta [/tex]

    when I use these, it doesn't work out.
     
  5. Sep 20, 2009 #4

    Redbelly98

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    Staff Emeritus
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    Homework Helper

    That correctly gave you x=68.8 m. If you use the analogous equation for Δy, you'll get y as well.

    Since the question asks for the location of the object, all you need is x and y. We don't need Vx, Vy, or the distance travelled.
     
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