Question

a ball is thrown directly upward from a height of 6 ft with an initial velocity of 24 ft/sec. the function ( s(t)=-16 t^{2}+24 t + 6 ) 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. the ball reaches its maximum height of ( square ) ft ( square ) sec(s) after the ball is thrown. (type integers or decimals.)

Explanation:

Step1: Find the time \(t\) at maximum height

For a quadratic function \(s(t)=at^{2}+bt + c\) (here \(a=-16\), \(b = 24\), \(c = 6\)), the time \(t\) at which the vertex (maximum for \(a<0\)) occurs is given by the formula \(t=-\frac{b}{2a}\).
Substitute \(a=-16\) and \(b = 24\) into the formula:
\(t=-\frac{24}{2\times(-16)}=\frac{24}{32}=\frac{3}{4}=0.75\)

Step2: Find the maximum height

Substitute \(t = 0.75\) into the function \(s(t)=-16t^{2}+24t + 6\)

$$
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$$

Answer:

The ball reaches its maximum height of \(15\) ft \(0.75\) sec(s) after the ball is thrown.