Answer
The magnitude of the initial velocity is $~~14.5~m/s$
Work Step by Step
Let "up" be the positive direction.
Let $v_0$ be the initial velocity of the ball.
Since the ball starts $43.3~m$ above the bottom of the lake, let $y_0 = 43.3~m$ and let the bottom of the lake be $0~m$
We can find the initial velocity:
$y-y_0 = v_0~t+\frac{1}{2}at^2$
$ v_0~t=y-y_0 -\frac{1}{2}at^2$
$ v_0=\frac{y-y_0 -\frac{1}{2}at^2}{t}$
$ v_0=\frac{0-43.3~m -\frac{1}{2}(-9.8~m/s^2)(4.80~s)^2}{4.80~s}$
$v_0 = 14.5~m/s$
The magnitude of the initial velocity is $~~14.5~m/s$
