Answer
The combined object travels a horizontal distance of 0.945 meters.
Work Step by Step
We can find the time $t$ to fall to the floor.
$y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{(2)(2.20~m)}{9.80~m/s^2}}$
$t = 0.670~s$
We can find the horizontal speed of the combined object when it leaves the table.
$m_2v_2= m_1v_1$
$v_2 = \frac{m_1v_1}{m_2}$
$v_2 = \frac{(0.500~kg)(24.0~m/s)}{8.50~kg}$
$v_2 = 1.41~m/s$
We can use the time and the horizontal speed to find the horizontal distance $x$
$x = v_2~t = (1.41~m/s)(0.670~s)$
$x = 0.945~m$
The combined object travels a horizontal distance of 0.945 meters.
