## College Physics (4th Edition)

(a) $h = L - L~cos~\theta$ (b) The speed at the bottom of the swing is $4.93~m/s$ (c) Jane can swing up to a maximum height of 1.24 meters.
(a) Let $L$ be the length of the vine. When the vine makes an angle of $\theta$ with the vertical, the vertical distance from the top of the vine is $L~cos~\theta$. Therefore, when the vine hangs straight down, the height difference $h = L - L~cos~\theta$ (b) We can assume that the gravitational potential energy at the bottom of the swing is zero. We can use conservation of energy to find the speed $v_2$ at the bottom of the swing. $\frac{1}{2}mv_2^2 = mgh_1+\frac{1}{2}mv_1^2$ $v_2^2 = 2gh_1+v_1^2$ $v_2 = \sqrt{2gh_1+v_1^2}$ $v_2 = \sqrt{(2)(9.80~m/s^2)(7.0~m)(1-cos~20^{\circ})+(4.0~m/s)^2}$ $v_2 = 4.93~m/s$ The speed at the bottom of the swing is $4.93~m/s$ (c) We can use conservation of energy to find the height $h_3$ that Jane can swing: $mgh_3 = \frac{1}{2}mv_2^2$ $h_3 = \frac{v_2^2}{2g}$ $h_3 = \frac{(4.93~m/s)^2}{(2)(9.80~m/s^2)}$ $h_3 = 1.24~m$ Jane can swing up to a maximum height of 1.24 meters.