## College Physics (4th Edition)

The speed at the bottom of the swing is $8.42~m/s$
We can find the difference in height between the starting point and the bottom of the swing: $h = (20.0~m)-(20.0~m)~cos~35.0^{\circ} = 3.62~m$ The kinetic energy at the bottom will be equal to the magnitude of the change in the potential energy: $\frac{1}{2}mv^2 = mgh$ $v = \sqrt{2gh}$ $v = \sqrt{(2)(9.80~m/s^2)(3.62~m)}$ $v = 8.42~m/s$ The speed at the bottom of the swing is $8.42~m/s$.