a stone is dropped into th water fro a bridge 44 m above the water. another stone is thrown vertically 1.0s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?
x = x0 + vx0t + 1/2 ax(t)
vx = vxo + axt
x = x0 + 1/2(vx+vx0)(t)
The Attempt at a Solution
it seemed like a pretty simple question, I calculated how long it took for a stone to go from 44 m above the water to 0 m in free fall ( 0 = 44 - 4.9(t) ) then I subtracted 1 second from that time and plugged it into 0 = 44 + vx(8) - 4.9(8) to see the initial velocity of the second stone..
then something strange happened.. I tried to plot a speed v.s. time graph and so I wanted to find the velocity of the stone at a point in time.. so I plugged vx = -4.9(9) and I get some pretty big number.. -88.. then I tried plugging it into x = 1/2(-88)(9) and I get a humongous number.. what's going on?