Question 608446
Malik popped a ball staight up with an initial upward velocity of 45 feet per second. the height h, in feet, of the ball above the ground is modeled by the equation h = 2 + 45t - 16t^2. How long was the ball in the air if the catcher catches the ball when it is 2 feet above the ground?
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t=seconds ball  is in the air after it is thrown
h=height ball is in the air after t seconds
h = 2 + 45t - 16t^2
given height is 2 ft
2 = 2 + 45t - 16t^2
0=45t-16t^2
16t^2-45t=0
t(16t-45)=0
t=0 (ball is 2 ft above the ground before it is thrown)
or 
16t-45=0
t=45/16≈2.81 sec (ball is 2 ft above the ground on its way down)