Physics
posted by Bland .
A ball is thrown directly upward with an initial velocity of 12 m/s. If the ball is released from an initial height of 2.9 m above ground, how long is the ball in the air before landing on the ground? Ignore air drag.

height above ground =
Y = 12 t  4.9 t^2 + 2.9
SWet that eq 
height above ground =
Y = 12 t 4.9 t^2 + 2.9 (meters)
Set Y equal to zero and solve for t.
4.9t^2 12t 2.9 = 0
There will be two solutions. Take the positive one.
Use thye "quadratic formula":
t = (1/9.8)[12 + sqrt(144 + 56.84)]
= 2.671 s