a ball is thrown directly upward from a height of 3 ft with an initial velocity of 24 ft/sec. the function…

Explanation:

Step1: Identify the function type

The height - function $s(t)=-16t^{2}+24t + 3$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 24$, and $c = 3$.

Step2: Find the time of maximum height

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate (in our case, the time $t$) of the vertex is given by $t=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 24$ into the formula: $t=-\frac{24}{2\times(-16)}=\frac{24}{32}=\frac{3}{4}=0.75$ seconds.

Step3: Find the maximum height

Substitute $t = 0.75$ into the function $s(t)=-16t^{2}+24t + 3$. $s(0.75)=-16\times(0.75)^{2}+24\times0.75 + 3$. First, calculate $(0.75)^{2}=0.5625$. Then $-16\times(0.75)^{2}=-16\times0.5625=-9$. And $24\times0.75 = 18$. So $s(0.75)=-9 + 18+3=12$ feet.

Answer:

The ball reaches its maximum height of 12 feet 0.75 seconds after the ball is thrown.