Posted by **Bland** on Tuesday, June 19, 2012 at 11:43pm.

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.

- Physics -
**drwls**, Wednesday, June 20, 2012 at 1:54am
height above ground =

Y = 12 t - 4.9 t^2 + 2.9

SWet that eq

- Physics -
**drwls**, Wednesday, June 20, 2012 at 2:02am
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

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