Asked by Anonymous
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?
Answers
Answered by
Henry
V^2 = Vo^2 + 2g*d
V^2 = 0 + 64*(876-200)
V^2 = 43,264
V = 208 Ft/s
V^2 = 0 + 64*(876-200)
V^2 = 43,264
V = 208 Ft/s
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