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Projectile motion question with angles and velocity

  1. Mar 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Joe Carter hit a fly ball during the spring training. It just cleared a 10m high vertical fence on the way down at 45 degrees as shown and struck the level ground 8m beyond the fence. Calculate the speed of the ball when it left his bat at ground level (start and end points are the same, ignore air resistance). Answer in m/s.

    2. Relevant equations

    ∆x = vit
    5 kinimatics equations for y direction

    3. The attempt at a solution

    Since air resistance is negligible, and the ball starts and ends at the same level, vi = vf
    a = 9.8 m/s down
    We can break equation into 2 parts - part 1 before fence and part 2 after
    For Part 2
    ∆dix = 8m
    ∆diy = 10m
    a = 9.8 m/s (down)
    vf = vi of part 1
    vf = ?

    It seems we're missing time and vf, two variables of the equation for part 2, and several variables for part 1. If you guys can help me figure this out that'd be sweet.
  2. jcsd
  3. Mar 15, 2008 #2
    Wherez the figure?
  4. Mar 15, 2008 #3
    Use coordinates with the spot where the ball cleared the fence as origin. The time coordinate will also be 0 when this happens. Call the speed of the ball at this point v.
    This is the initial speed.
    Since we know the angle we have v_x = cos(45)v and v_y = -sin(45)v.
    note that cos(45) = sin(45)

    Now write down expression for x and y as a function of t.

    x(t) = ....
    y(t) = ....

    The ball will hit the ground at x=8, y=-10. call the time this happens T.

    if you then set x(T) = 8 and y(T) =-10 you get 2 equations for v and T.

    The v you find from that is at the top of the fence, so you'll need to compute v_x(T) and v_y(T) to get the speed at the point the ball hits the ground.
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