#### Answer

The object's speed when it was released was 4.5 m/s

#### Work Step by Step

We can find the time $t$ to fall to the floor as:
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(1.0~m)}{9.80~m/s^2}}$
$t = 0.452~s$
We can find the horizontal speed $v_x$ when the object reached the edge of the table as:
$v_x = \frac{x}{t}$
$v_x = \frac{0.30~m}{0.452~s}$
$v_x = 0.664~m/s$
We can find the rate of deceleration as the object slid across the table as:
$F_f = ma$
$mg~\mu_k = ma$
$a = g~\mu_k$
$a = (9.80~m/s^2)(0.50)$
$a = 4.90~m/s^2$
We can find the speed when the object left the person's hand as;
$v_x^2 = v_{x0}^2+2ax$
$v_{x0}^2 = v_x^2-2ax$
$v_{x0} = \sqrt{v_x^2-2ax}$
$v_{x0} = \sqrt{(0.664~m/s)^2-(2)(-4.90~m/s^2)(2.0~m)}$
$v_{x0} = 4.5~m/s$
The object's speed when it was released was 4.5 m/s.