Question 1185585
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A ball is thrown directly upward from a height of 4 ft with an initial velocity of 24 ​ft/sec. 
The function ​s(t) gives the height of the​ ball, in​ feet, t seconds after it has been thrown. 
Determine the time at which the ball reaches its maximum height and find the maximum height.
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<pre>
The function s(t) is  s(t) = -16t^2 + 24t + 4.


The maximum height is reached when the quadratic function s(t) reaches its maximum.


It happens when  t = " -{{{b/(2a)}}} ", referring to the general form of a quadratic equation.


In your case a = -16, b = 24, THEREFORE,  {{{t[max]}}} = {{{-24/(-2*16)}}} = {{{24/32}}} = {{{3/4}}} of a second = 0.75 seconds.


The maximum height then is the height at t = {{{t[max]}}} = 0.75 seconds


    {{{h[max]}}} = -16*0.75^2 + 24*0.75 + 4 = 13 ft.


<U>ANSWER</U>.  {{{t[max]}}} = 0.75 seconds;  {{{h[max]}}} = 13 ft.
</pre>

Solved.


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