Answer
After 1.50 seconds, the clay is on the ground a horizontal distance of 26.4 meters away.
Work Step by Step
We can find the time it takes for the clay to drop $8.50~m$:
$\Delta y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2\Delta y}{a}}$
$t = \sqrt{\frac{(2)(8.50~m)}{9.80~m/s^2}}$
$ t= 1.32~m$
We can find the horizontal distance the clay travels in this time:
$\Delta x = v_x~t$
$\Delta x = (20.0~m/s)(1.32~s)$
$\Delta x = 26.4~m$
Since the clay sticks in place when it hits the ground, after 1.50 seconds, the clay is on the ground a horizontal distance of 26.4 meters away.
