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Relative Motion of a football player

  1. Jan 25, 2005 #1
    A football reciever B runs the slant in pattern, making a cut at P and thereafter running with a constant speed [tex]v_B=21ft/sec[/tex] at an angle of 30 degrees from the y-axis. The quarterback releases the ball with a Horizontal velocity [tex]v_A_x=100ft/sec[/tex] the instant the reciever passes pont P. Determine the angle at [tex]\alpha[/tex] (the angle between Va and the x-axis) and determine the velocity of the ball relative to the reciever when the ball is caught. Neglect any vertical motion of the ball.

    The answer is given [tex]\alpha=33.3[/tex] degrees and [tex]v_A_/_B= (73.1 i + 73.1 j)ft/sec[/tex]

    I tried to solve this by finding the time both the football and the reciever reach the same distance by using [tex]100*t = 21*sin(30)*t+45[/tex]
    I used this time to find the displacement of the reciever from (45,45)... then I tried to use [tex]tan(\frac{s_y}{s_x})=\alpha[/tex]. (the[tex]s_y[/tex] and [tex]s_x[/tex] I used were [tex]s_y=35.85[/tex] and [tex]s_x=50.28[/tex]But the alpha I get is 35.488 degrees, which is quite a bit higher. I was wondering what I am doing wrong.
    Thanks in advance

    Attached Files:

    Last edited: Jan 25, 2005
  2. jcsd
  3. Jan 26, 2005 #2
    You have two unknowns, [itex]\alpha[/itex] and t. You need two equations.

    If you set up your coordinates with point A at the origin, a vertical line through A as the y-axis, and the horizontal line through A as the x-axis, your expression
    21*sin(30)*t+45 gives the x-coordinate of the receiver at time t.
    but 100t gives the total distance the ball travels in that time along it's diagonal path. This doesn't help you. You need an expression for the x-coordinate of the ball at time t, i.e. the horizontal distance the ball travels in t seconds. Set that equal to your 21*sin(30)*t+45.

    But that's not enough. You need another equation equating the y-coordinates of the ball and the receiver at time t.

    Solve those 2 equations (using the appropriate trig identity) & you'll get your 33.3 degrees.
  4. Jan 26, 2005 #3
    Thanks, I figured out what I did wrong... The way the book worded the problem messed me up. They gave me the horizontal velocity, but the picture was top down... and basically I mistook the horizontal velocity for the x component of the throw, when it really was the magnitude.
    Thanks for your help.
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