## College Physics (4th Edition)

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.