Answer
$v_{0y} = 19.6~m/s$
Work Step by Step
In part (a), we found that the value of $v_{0x} = 10~m/s$
When $v_s = 0$, then $\Delta x_{bg} = 40~m$
Thus the time of flight is $4.0~s$
The time to reach maximum height is $t_1 = 2.0~s$
We can find $v_{0y}$:
$t_1 = \frac{v_{0y}}{g}$
$v_{0y} = g~t$
$v_{0y} = (9.8~m/s^2)(2.0~s)$
$v_{0y} = 19.6~m/s$
