SOLUTION: H(t)= -5t^2 + at + b At time t=0, a ball was thrown upward from the top of a building. Before the ball hit the ground, its height h(t), in feet, at the time t seconds is given by

Question 1088905: H(t)= -5t^2 + at + b
At time t=0, a ball was thrown upward from the top of a building. Before the ball hit the ground, its height h(t), in feet, at the time t seconds is given by the function above. A and b are constants. If the ball reached its maximum height of 125 feet at time t=3, what could be the height of the building?
Found 2 solutions by jorel1380, ikleyn:
Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
If h(t)=-5t�+at+b, then the maximum height is achieved at -a/2(-5) for t. So:
3=-a/-10
-a=-30
a=30
So, to find the height of the building, b, we have
125=-5(3�)+30(3)+b
125=90-45+b
b=80 ft. as the height of the building
☺☺☺☺
Answer by ikleyn(52786) (Show Source): You can put this solution on YOUR website!
.
The correct answer is 80 METERS, not 80 feet.

See my correct solution at this link
https://www.algebra.com/algebra/homework/Finance/Finance.faq.question.1088880.html

https://www.algebra.com/algebra/homework/Finance/Finance.faq.question.1088880.html

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