for each vertical motion model, identify the maximum height (in a. h(t)= - 16t² + 200t + 25 b. h(t)= - 16t²…

Explanation:

Step1: Recall vertex - formula for parabola

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate of the vertex is $t=-\frac{b}{2a}$, and the function value at the vertex gives the maximum (if $a<0$) or minimum (if $a > 0$) of the function. In the height - time function $h(t)=-16t^{2}+bt + c$, $a=-16<0$, so the function has a maximum.

Step2: Find the time $t$ at which maximum height occurs for part a

For $h(t)=-16t^{2}+200t + 25$, where $a=-16$ and $b = 200$. Using the formula $t=-\frac{b}{2a}$, we have $t=-\frac{200}{2\times(-16)}=\frac{200}{32}=\frac{25}{4}$.

Step3: Find the maximum height for part a

Substitute $t = \frac{25}{4}$ into $h(t)=-16t^{2}+200t + 25$. [ \begin{align*} h(\frac{25}{4})&=-16\times(\frac{25}{4})^{2}+200\times\frac{25}{4}+25\ &=-16\times\frac{625}{16}+1250 + 25\ &=-625+1250+25\ &=650 \end{align*} ]

Step4: Find the time $t$ at which maximum height occurs for part b

For $h(t)=-16t^{2}+36t + 4$, where $a=-16$ and $b = 36$. Using the formula $t=-\frac{b}{2a}$, we have $t=-\frac{36}{2\times(-16)}=\frac{36}{32}=\frac{9}{8}$.

Step5: Find the maximum height for part b

Substitute $t=\frac{9}{8}$ into $h(t)=-16t^{2}+36t + 4$. [ \begin{align*} h(\frac{9}{8})&=-16\times(\frac{9}{8})^{2}+36\times\frac{9}{8}+4\ &=-16\times\frac{81}{64}+\frac{324}{8}+4\ &=-\frac{81}{4}+\frac{324}{8}+4\ &=-\frac{162}{8}+\frac{324}{8}+\frac{32}{8}\ &=\frac{-162 + 324+32}{8}\ &=\frac{194}{8}=\frac{97}{4}=24.25 \end{align*} ]

Answer:

a. 650 b. 24.25