Abaseball is thrown straight up and travels to a height of 9.0 meters before returning back to the thrower. if the thrower catches the ball at the same height it was thrown, how long would the ball be in the air?

Explanation:

The trick to problems where something is thrown upwards, stops and falls back down is to  find half, going fromt he start to the top of its height, then doubling.

So, for this we use the equation vf = vi + at to solve for t, though we don't know vi, so for THAT we need vf^2-vi^2=2as.  So let's get started.

rearranging vf^2-vi^2=2as to solve for vi you get vi = sqrt(-(2as - vf^2))

plugging in the variables are

vf = 0 m/s

vi we are looking for

a = -9.8 m/s^2

s = 9 m

Then we want to solve vf = vi + at for t which is t = (vf-vi)/a.  Remember then, t is just from the bottom to the top, where it stops, so you multiply this by 2 to get the whole time in the air.  Let me know if you don't understand anything.

A side note, you could also use the equation s = vi*t + .5at^2 in the second step where solving for t would get two answers, the start and then when it falls back and is caught.  It's a little more complicated, but if you are comfortable with quadratics it could be faster.  

Again, let me know if you need any further explanation.  

its c i just checked it