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
To solve for the initial velocity (Vi), you need to take the square root of -156.8. However, since -156.8 is a negative number, the square root is not a real number. This means that there is no real solution for the initial velocity in this scenario.
This result suggests that there might be a mistake in your calculations or assumptions. Keep in mind that when solving for the initial velocity of a projectile, you need to consider the direction of the velocity as well.
To solve this problem correctly, you could try rechecking your calculations or providing more information about the rocket's motion, such as the maximum height it reached or the time it took to reach that height.