Question 732251
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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 = {{{-b/(2a)}}}.  In our case, the maximum is achieved at

    x = {{{-122/(2*(-16))}}} = {{{122/32}}} = 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).


<U>ANSWER</U>.  The maximum height of the ball is 932.6 feet.
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Solved.