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Kinematics in 2-dimesions

  1. Jan 30, 2008 #1
    [SOLVED] Kinematics in 2-dimesions

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

    A soccer player kicks the ball toward a goal that is 25.0 m in front of him. The ball leaves his foot at a speed of 19.2 m/s and an angle of 65.8 ° above the ground. Find the speed of the ball when the goalie catches it in front of the net. (Note: The answer is not 19.2 m/s.)

    2. Relevant equations



    V= sqrt(Vx^2+Vy^2)

    3. The attempt at a solution

    Ok let me define all the variables I've got so far

    V(0)= 19.2

    Vx0= 7.87
    x= 25 (12.5 for one half)
    gx= 0? (not so sure about this one)
    Vx= ?(at top of apex )

    Vy= 0 (at top of apex)
    y= 15.626
    Vy0= 17.5 (initial y veolicty)

    Ok so the first thing I did was find the inital velocity for the x and y componets using sine and cosine, so

    V0x= 19.2cos65.8= 7.87
    V0y=19.2sin65.8= 17.5

    I then used my inital y velocity to find out the heigth (y distance) of the shot

    Vy^2= Vy0^2+2gy
    0= (17.5)^2+2(-9.8)y
    y= 15.625

    I'm unsure where to go now in order to out the speed of the ball when the goalie catches it i know that V= sqrt (Vx^2+Vy^2). Should i try and figure out the two final velocities and then plug them into that equation?
    Last edited: Jan 30, 2008
  2. jcsd
  3. Jan 30, 2008 #2
    You need to find the time at which the ball reaches the position of the net. What you've written as [tex]g_x[/tex] is actually just acceleration in the x-direction, g is simply used to denote gravitational acceleration (which only occurs in whats defined as the y-direction in this problem). If air resistance is ignored, acceleration in the x-direction is 0.

    Thus you know speed in the x-direction will not change over the course of the motion. Once you've found the time, you need to find the x and y velocities, and as you said, the square root of the sum of their squares is the speed at that point.
  4. Jan 30, 2008 #3
    Ahhh i see it now, the height wasn't even needed in this problem, thanks for the tip.
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