## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Let $v_1$, $v_2$, and $v_3$ be the speed at the start of the 1.0 meter section, 2.0-meter section, and 3.0-meter section respectively. We can find $v_1$; $v_1^2= 0-2ax_1$ $v_1 = \sqrt{-2ax_1} = \sqrt{-(2)(-1.0~m/s^2)(1.0~m)}$ $v_1 = \sqrt{2.0}~m/s$ We can find $v_2$; $v_2^2= v_1^2-2ax_2$ $v_2 = \sqrt{v_1^2-2ax_2}$ $v_2 = \sqrt{(\sqrt{2.0}~m/s)^2-(2)(-6.0~m/s^2)(2.0~m)}$ $v_2 = \sqrt{26.0}~m/s$ We can find $v_3$; $v_3^2= v_2^2-2ax_3$ $v_3 = \sqrt{v_2^2-2ax_3}$ $v_3 = \sqrt{(\sqrt{26.0}~m/s)^2-(2)(-1.0~m/s^2)(3.0~m)}$ $v_3 = \sqrt{32.0}~m/s$ $v_3 = 5.66~m/s$ The smallest speed that the ball can leave the golf club is 5.66 m/s.