A batter hits a pitched ball when the center of the ball is 1.36 m above the ground.The ball leaves the bat at an angle of 45° with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 112 m. (a) Does the ball clear a 7.65-m-high fence that is 102 m horizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center? so i solved for the initial velocity which is 33.13m/s the i broke it up into its x and y components Vx=23.43 m/s Vy = 23.43 m/s since the range is 112m i found the time that it takes for the ball to go 112m t= x/Vx = 112m/23.43m/s i then found the time it takes to 102 m using the same method and got 4.35s so to find the height of the ball i did x=(23.43)(4.35)+1/2(9.8)(4.35)^2 = 9.2m so i thought that it would be 2.91m above the fence but thats wrong. where did i go wrong/?