SOLUTION: For problems 18 through 20, the height, h, of a ball t seconds after it is released for a jump shot in a basketball game can be modeled with the equation {{{ h=-16t^2+12t+8 }}} .
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-> SOLUTION: For problems 18 through 20, the height, h, of a ball t seconds after it is released for a jump shot in a basketball game can be modeled with the equation {{{ h=-16t^2+12t+8 }}} .
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Question 189481: For problems 18 through 20, the height, h, of a ball t seconds after it is released for a jump shot in a basketball game can be modeled with the equation .
19. State the number of solutions to the problem and state if the ball has a chance of going into the basket.
a. There are 2 solutions, meaning 2 times when the ball is 10 ft. high. The time that the ball reaches 10 feet on its way up and on its way down. One of these times will be the ball's opportunity to pass through the hoop.
b. The equation does not allow two solutions to the problem. This ball will not reach ten feet.
c. The ball reaches ten feet as its maximum height so there is only one solution. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! For problems 18 through 20, the height, h, of a ball t seconds after it is released for a jump shot in a basketball game can be modeled with the equation
h = -16t^2 + 12t + 8 .
:
19. State the number of solutions to the problem and state if the ball has a chance of going into the basket.
:
a. There are 2 solutions, meaning 2 times when the ball is 10 ft. high. The time that the ball reaches 10 feet on its way up and on its way down. One of these times will be the ball's opportunity to pass through the hoop.
-16t^2 + 12t + 8 = 10
-16t^2 + 12t + 8 - 10 = 0
-16t^2 + 12t - 2 = 0
Simplify and change the signs, Divide by -2
8t^2 - 6t + 1 = 0
Factor
(4t - 1)(2t - 1) = 0
Two solutions:
4t = 1
t = sec
and
2t = 1
t = sec; could just barely pass thru the hoop
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