## University Physics with Modern Physics (14th Edition)

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.