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Homework Help: Kinematics in Two Dimensions

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data
    A soccer player kicks the ball toward a goal that is 16.8 m in front of him. The ball leaves his foot at a speed of 16.0 m/s and an angle of 28.0 degrees above the ground. Find the speed of the ball when the goalie catches it in front of the net.

    2. Relevant equations
    x = vt
    x = Vot+(1/2)at2

    3. The attempt at a solution
    Vx = (cos 28)(16.0) = 14.1 m/s
    Vy = (sin 28) (16.0) = 7.5 m/s

    x = 16.8 m
    V = 14.1 m/s

    a = -9.8 m/s2
    Vyo = 0 m/s
    (V = 7.5 m/s?)

    I'm stuck as to where I should go from there or even if I started out correctly.
    If you can assist me, I would be very grateful. Thank you!
  2. jcsd
  3. Oct 4, 2008 #2
    Yes, you have started well. First you want to find how long the ball is in the air for.
    There is no acceleration in the x plane so the x speed will remain the same.
    So you can find the time using only the x parameters.

    However the final speed of the ball will depend on x and y velocities.
    You will now have:


    use [tex] v_{final} = v_{initial} + at [/tex]

    solve for v final y component.
    Then use pythagoras to find total speed (combining x and y components).
  4. Oct 4, 2008 #3
    Thank you for replying.

    I'm still a bit puzzled though, but if I understood correctly, this is what I did:

    x = vt
    (16.8 m) = (14.1 m/s)t
    t=16.8 m/14.1
    t= 1.2 s

    Vf2 = Vo + at
    Vf = (7.5 m/s) + (-9.8 m/s2)(1.2 s)
    Vf = -4.3 m/s

    I don't understand what to use pythagoreas for though.

    Please bear with me. I'm not very good at word problems :(
  5. Oct 5, 2008 #4
    Well now you have y velocity and the x velocity so you want to combine them to total velocity.

    [tex] \sqrt{-4.3^2 + 14.1^2} [/tex]

    it makes a right angled triangle when you think about it, because x and y are perpendicular to each other, so the line that joins them (total velocity) is the hypotenuse.
  6. Oct 5, 2008 #5
    Oh ok, yeah it is the hypotenuse.

    Thank you for your help!
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