Answer
$v_0 = 24~m/s$
Work Step by Step
We can find $v_{0y}$:
$y = y_0+v_{0y}~t+\frac{1}{2}a_y~t^2$
$0 = y_0+v_{0y}~t+\frac{1}{2}a_y~t^2$
$v_{0y}~t = -y_0-\frac{1}{2}a_y~t^2$
$v_{0y} = -\frac{y_0}{t}-\frac{1}{2}a_y~t$
$v_{0y} = -\frac{1.5~m}{4.5~s}-\frac{1}{2}(-9.8~m/s^2)~(4.5~s)$
$v_{0y} = 21.72~m/s$
We can find $v_x$:
$x = v_x~t$
$v_x = \frac{x}{t}$
$v_x = \frac{46~m}{4.5~s}$
$v_x = 10.22~m/s$
We can find the magnitude of the initial velocity:
$v_0 = \sqrt{(10.22~m/s)^2+(21.72~m/s)^2} = 24~m/s$
