SOLUTION: height y (in feet) of punted football is approximated by y = -16/2025 x^2 +9/5x + 5/2 where x is horizontal distance (in feet ) from point at which the ball is punted .
(a)use a
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-> SOLUTION: height y (in feet) of punted football is approximated by y = -16/2025 x^2 +9/5x + 5/2 where x is horizontal distance (in feet ) from point at which the ball is punted .
(a)use a
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Question 1058616: height y (in feet) of punted football is approximated by y = -16/2025 x^2 +9/5x + 5/2 where x is horizontal distance (in feet ) from point at which the ball is punted .
(a)use a graphing utility to graph the path of football.
(b)how high is the ball when it is punted ?(find y when x=0)
(c)what is the maximum height of football?(round your answer to two decimal places)
(d)how far from punter does the football strike the ground?(round your answer to two decimal places) Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! height y (in feet) of punted football is approximated by y = -16/2025 x^2 +9/5x + 5/2 where x is horizontal distance (in feet ) from point at which the ball is punted.
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(a)use a graphing utility to graph the path of football.
dl the FREE graph software at www.padowan.dk
Use Insert, enter -16x^2/2025 + 9x/5 + 5/2
Pick a color
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(b)how high is the ball when it is punted ?(find y when x=0)
2.5 feet
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(c)what is the maximum height of football?(round your answer to two decimal places)
It's the vertex of the parabola at x = -b/2a
x = (-9/5)/(-32/2025) = 9*2025/150 = 121.5
y = 31.66 feet
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(d)how far from punter does the football strike the ground?(round your answer to two decimal places)
Find x when y = 0
y = -16/2025 x^2 +9/5x + 5/2 = 0
16x^2 - 3645x - 5062.5 = 0
