a bungee cord jumper jumps off the new river bridge. the bridge is 876 feet high. How fast is the jumper falling when he reaches 200 feet above the ground?

Answer 1

V^2 = Vo^2 + 2g*d
V^2 = 0 + 64*(876-200)
V^2 = 43,264
V = 208 Ft/s