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Projectile Motion of football

  1. Feb 13, 2006 #1
    A football is kicked at ground level with a speed of 40.0 m/s at an angle of 39.0° to the horizontal. How much later does it hit the ground?

    X0= 0m Y0= 0m
    X= ? Y= 0 m
    Vx=? Vy= ?
    Ax=0 m/s/s Ax=-9.8 m/s/s
    T=?
    V= 40.0 m/s
    Theta= 39.0

    Vx= 40.0 Cos (39.0) = 10.67 m/s
    Vy= 40.0 Sin (39.0)= 38.55 m/s

    0= 38.55T - 4.9 T^2

    -38.55 +/- sqrt ((38.55)^2-4(-4.9)(0))/ 2(-4.9)
    -38.55 +/- 38.55/-9.8
    T= 0 and 7.87 seconds <-------------------------------I dont belive this is right. Could someone help me determine where I wondered astray?
     
  2. jcsd
  3. Feb 13, 2006 #2
    The 10.67 is incorrect.
    40 * cos(39) = 31.1

    40*sin(39) = 25.2

    The "structure" and signs of this formula are correct though

    marlon
     
  4. Feb 14, 2006 #3
    Thanks! Just forgot to swithc the calculater from radians to degrees
     
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