## College Physics (4th Edition)

The maximum height of the swing is $~2.3~meters~$ above the ground.
We can use conservation of energy to find the height $h_f$ when the swing reaches its maximum height: $U_f+KE_f= U_0+KE_0$ $mgh_f+0 = mgh_0+\frac{1}{2}mv_0^2$ $h_f = h_0+\frac{v_0^2}{2g}$ $h_f = (0.5~m)+\frac{(6.0~m/s)^2}{(2)(9.80~m/s^2)}$ $h_f = 2.3~m$ The maximum height of the swing is $~2.3~meters~$ above the ground.