SOLUTION: A ball is tossed upward with an initial velocity of 122 ft/s from a platform that is 700 ft above the surface of the earth. After t seconds, the height of the ball above the ground
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Question 732251: A ball is tossed upward with an initial velocity of 122 ft/s from a platform that is 700 ft above the surface of the earth. After t seconds, the height of the ball above the ground is given by the equation h = -16t^2 + 122t + 700. What is the maximum height of the ball? Round to the nearest tenth of a foot. Found 2 solutions by lynnlo, ikleyn:Answer by lynnlo(4176) (Show Source):
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A ball is tossed upward with an initial velocity of 122 ft/s from a platform that is 700 ft above
the surface of the earth. After t seconds, the height of the ball above the ground is given
by the equation h = -16t^2 + 122t + 700. What is the maximum height of the ball?
Round to the nearest tenth of a foot.
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This given equation h = -16t^2 + 122t + 700 has the leading coefficient negative, -16.
So, it describes a parabola opened downward. Such a parabola has a maximum.
According to the general theory, a parabola y = ax^2 + bx + c with a negative leading coefficient 'a'
has a maximum at the point x = . In our case, the maximum is achieved at
x = = = 3.8125.
It means that the maximum height is achieved at 3.8125 second after tossing.
To find the maximum height h, substitute x = 3.8125 into the formula and calculate
h = -16*3.8125^2 + 122*3.8125 + 700 = 932.6 feet (rounded as requested).
ANSWER. The maximum height of the ball is 932.6 feet.
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