Question 1203130
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A ball is thrown straight upward with a velocity of 90 ft/sec. 
Its height above the ground after t seconds is h = 90t - 16t2.
What is the maximum height that the ball reaches?
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<pre>
This quadratic function can be factored into the product

    h = 90t - 16t^2 = t*(90-16t).


From this decomposition, the zeroes of this quadratic function (its x-intercepts) are

    t = 0 second and  t = {{{90/16}}} = 5.625 seconds.


This quadratic function achieves its maximum value half-way between x-intercepts.

So,  {{{t[max]}}} = {{{5.625/2}}} = 2.8125 seconds.


To find the maximum height, simply substitute this value  t = {{{t[max]}}} = 2.8125  into the formula and get

    {{{h[max]}}} = 90*2.8125 - 16*2.8125^2 = 126.5625 ft = 126.6 ft  (rounded as requested).
</pre>

Solved.