Physics
posted by Anonymous .
You shoot a rocket straight into the air from the ground.It takes 8.0 seconds to come back down to the ground.What was its initial velocity ?
Vf^2=Vi^2+2ax
0=vi^2+2(9.8)(8.0) < My work
vi^2=156.8
I don't how to solve the last step

You could solve it by taking the square root of both sides of the equation. However, your equation is wrong. You are apparently trying to apply energy conservation at the top of the trajectory, where Vf = 0. In that case, the "x" term is not 8.0 (seconds). It should be the height of the trajectory, in meters.
The rocket spends 4.0 of the 8.0 seconds going up. Decelerating at a rate g, we have
g*t = Vf = 9.8*4 = 39.2 m/s
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