#### Answer

The launch speed is 7.4 m/s

#### Work Step by Step

We can find the time it takes for the steel ball to fall 2.5 meters.
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(2.5~m)}{9.80~m/s^2}}$
$t = 0.714~s$
We can find the horizontal component of the plastic ball's speed.
$v_x = \frac{x}{t} = \frac{4.0~m}{0.714~s}$
$v_x = 5.6~m/s$
We can find the vertical component of the plastic ball's launch speed.
$y = v_{0y}t-\frac{1}{2}gt^2$
$v_{0y} = \frac{y+\frac{1}{2}gt^2}{t}$
$v_{0y} = \frac{1.0~m+\frac{1}{2}(9.80~m/s^2)(0.714~s)^2}{0.714~s}$
$v_{0y} = 4.9~m/s$
We can find the launch speed.
$v = \sqrt{(v_x)^2+(v_y)^2}$
$v = \sqrt{(5.6~m/s)^2+(4.9~m/s)^2}$
$v = 7.4~m/s$
The launch speed is 7.4 m/s