SOLUTION: The height of a ball thrown upward from the top of a building is given by: h(t)=-16t^2+64t+80
What is the height of the ball at 1 second?
What is the maximum height reached b
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-> SOLUTION: The height of a ball thrown upward from the top of a building is given by: h(t)=-16t^2+64t+80
What is the height of the ball at 1 second?
What is the maximum height reached b
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Question 1042039: The height of a ball thrown upward from the top of a building is given by: h(t)=-16t^2+64t+80
What is the height of the ball at 1 second?
What is the maximum height reached by the ball?
How long will it take for the ball to reach the ground? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The height of a ball thrown upward from the top of a building is given by: h(t)=-16t^2+64t+80
What is the height of the ball at 1 second?
h(1) = -16+64+80 = 128 ft.
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What is the maximum height reached by the ball?
Max occurs when x = -b/(2a) = -64/(2*-16) = 2 sec
h(2) = -16(2^2) + 64(2) + 80 = 144 ft
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How long will it take for the ball to reach the ground?
When it reaches the ground the height is zero.
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Solve:: -16t^2+64t+80 = 0
t^2 - 4t - 5 = 0
(t-5)(t+1) = 0
Positive solution::
t = 5 seconds
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Cheers,
Stan H.
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