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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
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)
To get the positive solution:
t = [-44 - sqrt(74064)]/(-32)
t = [-44-272.15]/(-32)
t = 9.88 seconds
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
C. How long does it take the ball to reach its maximum height? Answer is 1.37 seconds.