## College Physics (4th Edition)

We can use conservation of energy to find the skier's speed at the bottom of the slope: $\frac{1}{2}mv^2 = mgh$ $v = \sqrt{2gh}$ $v = \sqrt{(2)(9.80~m/s^2)(5.0~m)}$ $v = 9.90~m/s$ We can use conservation of momentum to find the speed after picking up the backpack: $m_f~v_f = m_0~v_0$ $v_f = \frac{m_0~v_0}{m_f}$ $v_f = \frac{(65~kg)(9.90~m/s)}{65~kg+20~kg}$ $v_f = 7.63~m/s$ We can find the time it takes to fall 2.0 meters: $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(2.0~m)}{9.80~m/s^2}}$ $t = 0.639~s$ We can find the horizontal distance the skier (with the backpack) moves in this time: $x = v_x~t = (7.63~m/s)(0.639~s) = 4.9~m$ The man lands a horizontal distance of 4.9 meters from the edge of the ledge.