1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

2D Motion

  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 50.0 m/s at an angle of 40.0 degrees above the horizontal. What is the range of the object?

    2. Relevant equations
    All the 2D Motion equations - too many to write all out. Ex:

    [tex] V_x = V_0x + a_xt [/tex]


    3. The attempt at a solution
    I don't even think there is enough information given to solve this. I can only solve for
    [tex] V_x [/tex] which is [tex] 50 * cos 40 = 38.3 [/tex] from there I don't know where to go. I try to figure out the y components of this motion, but we don't really know anything about y. We can't really find [tex] V_y [/tex] because to find that we need the equation [tex] V_y = v_0 sin \theta - gt [/tex] but we don't know the time.
     
  2. jcsd
  3. Sep 20, 2009 #2
    To work out the range we need to work out the flight time. So first find the time it takes the ball to reach it's highest point using the y-component of it's velocity. Then calculate how long it takes to hit the floor. The sum of these two times is the flight time.
     
  4. Sep 20, 2009 #3
    ARRRGGGHHHHH - silly me. How could I miss that?

    Thank you for pointing me in the right direction, much appreciated m8.
     
  5. Sep 20, 2009 #4
    Np :)
     
  6. Sep 20, 2009 #5
    An simpler, more fruitful approach would be to develop two equations, and then turn them into one.

    One would be [tex]x(t)[/tex], the other would be [tex]y(t)[/tex] and combining them will give you [tex]y(x)[/tex]

    That function will describe the height of the object above the the origin as a function of its x-axis distance from the origin. Once you have that function, you can just set [tex]y=0[/tex] to find the x for which this holds true (The range.)

    To get you started, I'll just rewrite your equation:

    [tex]x(t)=v_0\cos{\theta}\cdot t[/tex]

    [tex]t=\frac{x}{v_0\cos{\theta}}[/tex]

    From here, you should be all set.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: 2D Motion
  1. Motion in 2D (Replies: 1)

  2. 2d motion (Replies: 1)

  3. 2d motion (Replies: 2)

  4. 2D motion (Replies: 9)

  5. 2D Motion (Replies: 6)

Loading...