Numbers gone wrong, free fall

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

x = x0 + vx0t + 1/2 ax(t)
vx = vxo + axt
x = x0 + 1/2(vx+vx0)(t)

3. 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?

thanks

 

I'm not sure :S, that equation should work shouldn't it? I've used it before.. Also, my x0 is 44 because I'm taking that as the initial distance and so my acceleration is negative because it will go from 44m to 0m


thanks

 

the equation is x = x0 + vx0t (initial velocity) + 0.5at^2 and it should work, as it comes from x = x0 + 1/2(v + vx0 )t which is just the addition of distances..

thanks

 

can someone help please..

 

AH right.. I forgot that it was at^2

thanks a lot!!