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Soccer Ball Projectile Prblm

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A soccer ball is kicked at an angle of 15 degrees with the ball traveling at 15.0 m/s
    a) what is the max height reached by the ball?
    b) max range?
    c) how could range be increased?

    so i need to find:
    ymax
    R = xmax

    2. Relevant equations
    these i assume are the given:
    [tex]\vartheta[/tex]=15 degrees
    v0=15.0 m/s
    ay= -g
    x0, y0 and y = 0


    3. The attempt at a solution
    vx0 = v0 cos 15 = 15 cos 15 = 14.5
    vy0 = v0 sin 15 = 15 sin 15 = 3.88

    vy = 0 = vy - gtu <---tu is time of ball movin upward

    must get tu:

    = vy0 / (g)

    here is where i get stuck cuz i dont know what to do with the g since i dont have a value for it.
     
  2. jcsd
  3. Feb 3, 2009 #2
    g is acceleration due to gravity. It is about 9.8 m/s^2
     
  4. Feb 3, 2009 #3
    so g is always 9.8?
    is that just a given?

    what do i do next?
     
  5. Feb 3, 2009 #4
    yeah. i think sometimes g is rounded to 10 or 9.81. i think so unless the problem states otherwise. So use that to find the time.

    If you want the max height, you should probably use this kinematic equation:

    vyf^2 = vyi^2 + 2at

    Note: a = -g in this case

    For the range, you use d = (vx) (t) because there is no acceleration in the horizontal
     
  6. Feb 3, 2009 #5
    max range occurs at angle 45 degrees
     
  7. Feb 4, 2009 #6
    so i got for the max height: 0.772 m <--thats about 3 ft. sounds right for a soccer ball kicked at 15 degrees?

    and max range to be: 11.5 m
     
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