Calculus
posted by Justyna on .
A rock is thrown upward off a bridge. In 2sec. the rock passes you on the way down. In 3 more sec. it hits the water. How high is the bridge above the water and how high did the rock go above the bridge?
Can someone help me start his problem!?

A rock is thrown upward off a bridge. In 2sec. the rock passes you on the way down. In 3 more sec. it hits the water. How high is the bridge above the water and how high did the rock go above the bridge?
If it passes you on the way down in 2 seconds, it took 1 second to reach its peak height.
Vf = Vo  32t = 0 = Vo  32(1) making the initial upward velocity 32fps and the velocity when passing you the same 32fps.
The distance from you to the water is derived from d = Vot + 32t^2 = 32(3) + 16(3)^2 = 96 + 144 = 240 ft.