## College Physics (4th Edition)

The speed at point 1 is $24.8~m/s$ The speed at point 2 is $18.0~m/s$ The speed at point 3 is $20.5~m/s$
We can use conservation of energy to find the speed at each point. At point 1: $mgh_1+\frac{1}{2}mv_1^2 = mgh_4+\frac{1}{2}mv_4^2$ $v_1^2 = 2gh_4-2gh_1+v_4^2$ $v_1 = \sqrt{2g~(h_4-h_1)+v_4^2}$ $v_1 = \sqrt{(2)(9.80~m/s^2)(20.0~m-0)+(15~m/s)^2}$ $v_1 = 24.8~m/s$ The speed at point 1 is $24.8~m/s$ At point 2: $mgh_2+\frac{1}{2}mv_2^2 = mgh_4+\frac{1}{2}mv_4^2$ $v_2^2 = 2gh_4-2gh_2+v_4^2$ $v_2 = \sqrt{2g~(h_4-h_2)+v_4^2}$ $v_2 = \sqrt{(2)(9.80~m/s^2)(20.0~m-15.0~m)+(15~m/s)^2}$ $v_2 = 18.0~m/s$ The speed at point 2 is $18.0~m/s$ At point 3: $mgh_3+\frac{1}{2}mv_3^2 = mgh_4+\frac{1}{2}mv_4^2$ $v_3^2 = 2gh_4-2gh_3+v_4^2$ $v_3 = \sqrt{2g~(h_4-h_3)+v_4^2}$ $v_3 = \sqrt{(2)(9.80~m/s^2)(20.0~m-10.0~m)+(15~m/s)^2}$ $v_3 = 20.5~m/s$ The speed at point 3 is $20.5~m/s$.