SOLUTION: 82. sports- You throw a baseball straight up into the air a 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

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Question 179588: 82. sports- You throw a baseball straight up into the air a 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.
b. find the height of the baseball after 0.5 second.
c. How many seconds is the baseball in the air (use a graphing utility to verify the answer.)

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
You throw a baseball straight up into the air a 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.
h(t) = -16t^2 + 45t + 5.5
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b. find the height of the baseball after 0.5 second.
h(0.5) = -16(0.5)^2 + 45(0.5) + 5.5 = 24 ft.
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c. How many seconds is the baseball in the air (use a graphing utility to verify the answer.)
-16t^2 + 45t + 5.5 = 0
Use the quadratic formula to get:
t = [-45 +- sqrt(45^2 - 4*-16*5.5)]/-32
Positive solution:
t = [-45 - sqrt(2377)]/-32
t = 2.93 seconds
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Cheers,
Stan H.