Answer
The ball clears the wall by a vertical distance of $~~4.31~m$
Work Step by Step
We can find the height $y$ at $t = 5.00~s$:
$y = y_0+v_{0y}~t+\frac{1}{2}a_y~t^2$
$y = (0.762~m)+(33.53~m/s)~(sin~55.0^{\circ})(5.00~s)+\frac{1}{2}(-9.8~m/s^2)~(5.00~s)^2$
$y = 15.59~m$
We can find the height above the top of the wall:
$15.59~m - 11.28~m = 4.31~m$
The ball clears the wall by a vertical distance of $~~4.31~m$
