The x-axis is horizontal and directed toward the point where the ball crosses over the wall. The time t required for the ball to reach the wall is given by the following equation in terms of the horizontal distance Ax = d = 123 m, the initial velocity Vo and the ball's horizontal component Vox* Step 2 Ax Ax t = Vox cos e 123 123 m Ax Vo cos 34 143 148 m = 0.86 0.829 Vo At this time, the ball must be a distance h = 11 m above the ground, or Ay = h - 1.0 m above its launch point. Using the equation for displacement under constant vertical accelerationAy = Voyt + we found above, we obtain the following equation. Remember that the vertical acceleration is the acceleration due to gravity, -g = 9.80 m/s. Step 3 148 Ay = (V sin e) 2 148 m 148 148 m + 2 (-9.80 m/s) Vo at and substituting the expression for t that For the vertical distance where the ball clears the wall, we havey = h - 1.0 m = (11 - 1.0) m = 10 m.Substituting this value for Ay, simplifying the equation we found in the last step, and solving for the initial velocity gives (-9.80 m/s) 148 Vo = 2(10 10 m - 148 2 148 m 38.40 38.4 m/s. 148 m sin 34 Step 4 (b) Now we substitute the value from the previous step for the initial velocity of the ball in the expression for t. The elapsed time when the ball reaches the wall is Submit t = S. m Skip (you cannot come back). m m/s