## College Physics (4th Edition)

When the sandbag is released, it will have an initial velocity of $10.0~m/s$ upward. We can find the velocity when the sandbag hits the ground: $v_f^2 = v_0^2+2a\Delta y$ $v_f = \sqrt{v_0^2+2a\Delta y}$ $v_f = \sqrt{(10.0~m/s)^2+(2)(-9.80~m/s^2)(-40.8~m)}$ $v_f = -30.0~m/s$ The sandbag's speed when it hits the ground is 30.0 m/s