Question 189179
You throw a baseball straight up into the air at a velocity of 45 feet per second. You release the baseball at a height of 5.5 feet and catch it when it falls back to a height of 6 feet. 
a. Use the position equation to write a mathematical model for the height of the baseball. 
y = -16t^2 + 45t + 5.5 at any time t.  y in feet, t in seconds
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b. Find the height of the baseball after 0.5 seconds.
Sub 0.5 into the equation
y = -16*0.25 + 45*0.5 + 5.5
y = -4 + 22.5 + 5.5
y @ 0.5 = 24 feet
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c. How many seconds is the baseball in the air? Use a graphing utility to verify your answer.
Find the times when the ball is at 6 feet.  There are 2 of them, 1 going up, 1 coming down.
6 = -16t^2 + 45t + 5.5
16t^2 - 45t + 0.5 = 0
t = (45/32) - sqrt(1993)/32 (going up) and
t = (45/32) + sqrt(1993)/32 (coming down)
t = ~ 2.801 seconds
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The graph looks good.