Answer
$v_0 = 31.9~m/s$
Work Step by Step
We can find $v_x$:
$v_x = \frac{50.0~m}{4.00~s} = 12.5~m/s$
By symmetry, the ball takes $1.00~s$ to fall from the top of the wall to the point where the ball is caught.
The total time of flight is $6.00~s$
The time to reach maximum height is $t_1 = 3.00~s$
We can find $v_{y0}$:
$t_1 = \frac{v_{y0}}{g}$
$v_{y0} = g~t_1$
$v_{y0} = (9.8~m/s^2)(3.00~s)$
$v_{y0} = 29.4~m/s$
We can find the magnitude of the ball's initial velocity:
$v_0 = \sqrt{(12.5~m/s)^2+(29.4~m/s)^2}$
$v_0 = 31.9~m/s$
