Question 397241: This is a quadratic function question but I didn't see that as an option to select.
A ball is tossed straight up from the top of a 1,127 foot tall building with an initial speed of 44 feet per second. The height h(t) of the ball t seconds after it is thrown is given by h(t)= -16t^2 + at + b (t^2 is my way of saying t-squared), where a is the initial speed of the ball and b is the initial height from which it is thrown.
A. How long was the ball in the air before it hit the street below? The answer is 9.88 seconds but I can't figure out how to get to the answer. I set it up as follows:
h(t)= -16t^2 + 44t + 1127
B. What is the maximum height the ball reaches? Answer is 1157.25 feet- again not sure how to get this.
C. How long does it take the ball to reach its maximum height? Answer is 1.37 seconds.
Even if you can just help with letter A. that would be a huge help.
Thank You
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A. How long was the ball in the air before it hit the street below? The answer is 9.88 seconds but I can't figure out how to get to the answer. I set it up as follows:
h(t)= -16t^2 + 44t + 1127
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h(t) is the height of the ball after t seconds.
The height is zero when the ball is back on the ground.
Solve: -16t^2 + 44t + 1127 = 0
Use the quadratic formula to get:
t = [-44 +- sqrt(44^2-4*-16*1127)]/(-32)
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To get the positive solution:
t = [-44 - sqrt(74064)]/(-32)
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t = [-44-272.15]/(-32)
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t = 9.88 seconds
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B. What is the maximum height the ball reaches? Answer is 1157.25 feet- again not sure how to get this.
Max height occurs when x = -b/(2a) = -44/(2*-16) = 1.375 seconds
Max height at that time is h(1.375)
h(1.375) = -16(1.375)^2+44(1.375)+1127 = 1157.3 feet
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C. How long does it take the ball to reach its maximum height? Answer is 1.37 seconds.
Answered above.
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Cheers,
Stan H.
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