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A woman throws a ball at a vertical wall d = 5.0 m away.

  1. Feb 11, 2015 #1
    A woman throws a ball at a vertical wall d = 3.2 m away. The ball is h = 1.6 m above ground when it leaves the woman's hand with an initial velocity of 14 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged. (Ignore any effects due to air resistance.)

    (a) Where does the ball hit the ground?
    m (away from the wall)

    (b) How long was the ball in the air before it hit the wall?

    (c) Where did the ball hit the wall?
    m (above the ground)

    (d) How long was the ball in the air after it left the wall?

    I found part B was .323s and part C is 4.29m. I don't know how to solve the second part.
  2. jcsd
  3. Feb 11, 2015 #2


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    You should have calculated the horizontal and vertical components of the ball's velocity when it struck the wall. Use the hint at the end of the problem statement and figure out the trajectory of the ball after it bounces off the wall. (bold portion).
  4. Feb 11, 2015 #3
    After the ball hits the wall, you have to start a new problem with an initial velocity and height again (except going in the opposite direction).

  5. Feb 11, 2015 #4
    I found the initial velocity in the x direction to be 14cos(225)=5.14 m/s and the initial velocity in the y direction to be 14sin(225)=-13.02 m/s. The equation I am trying to plug it into is V1y(SQ)=V0y(SQ)+2aΔt
  6. Feb 11, 2015 #5
    sorry meant Δy at the end of that equation
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