Answer
The minimum magnitude of the initial velocity is $~~v = 20~m/s$
Work Step by Step
The ball begins a vertical distance of $0.76~m$ above the top of the net.
We can find the time it takes for the ball to fall a vertical distance of $0.76~m$:
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(0.76~m)}{9.8~m/s^2}}$
$t = 0.394~s$
The ball must reach the net within this time.
We can find the minimum magnitude of the horizontal velocity:
$v_x = \frac{8.0~m}{0.394~s}$
$v_x = 20~m/s$
Since the initial velocity of the ball is horizontal, the minimum magnitude of the initial velocity is $~~v = 20~m/s$
