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Parabolic Projectile Motion problem.

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s at an angle of 40 degrees above the horizontal. It rises to it's maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

    So I say that we have these variables:
    Yi(Initial y position) = 0
    Yf(Final y position) = 3m
    Xi(Initial x position) = 0
    Xf(Final x position) = unknown
    Vi(Initial velocity) = 14
    Ti(Initial time) = 0
    AYc(Gravity y accel) = -9.8
    Initial angle of motion = 40 deg

    2. Relevant equations
    I think these equations are relevant
    Horizontal Velocity = magnitude * Cosine(theta)
    Vertical Velocity = magnitude * sin (theta)
    Xf = Xi + VXi(Tf - Ti) + 1/2 Ax (Tf - Ti)^2
    Yf = Yi + VYi(Tf - Ti) + 1/2 Ax (Tf - Ti)^2


    3. The attempt at a solution
    I drew a crappy diagram to get the vision of the green being higher than the tee.
    My XY axis cross at ground level.

    First I tried getting the direction specific velocities based on the angle and the initial velocity
    VYi = 14 sin(40) == 8.999
    VXi = 14 cos(40 == 10.725

    I guessed that I would substitute those into the 3rd/4th equations listed up top. But I didn't know Ax so I assume it is constant? I am so confused about where to go from here.

    Can anyone point me in the right direction?

    Thanks,
    Cory
     
  2. jcsd
  3. Feb 9, 2012 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    The whole idea behind projectile motion is that it is motion of an object that is in free fall, meaning that it is under the influence of gravity only. No other forces are acting. As a result, Newton's 2nd law says that a_x = 0. There is no horizontal acceleration. That simplifies your x equation somewhat.
     
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