question 4 incorrect 2 tries left. please try again. two rocks are launched straight up in the air. the…

Explanation:

Step1: Recall vertex - formula for a quadratic function

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate of the vertex (which gives the time at which maximum height is reached for height - time functions) is $t=-\frac{b}{2a}$, and the $y$ - coordinate (maximum height) is $y = a(-\frac{b}{2a})^{2}+b(-\frac{b}{2a})+c$. For $f(t)=4 + 30t-16t^{2}$, $a=-16$, $b = 30$, $c = 4$.

Step2: Calculate the time of maximum height for Rock A

$t_A=-\frac{b}{2a}=-\frac{30}{2\times(-16)}=\frac{30}{32}=\frac{15}{16}$ seconds.

Step3: Calculate the maximum height of Rock A

$f(\frac{15}{16})=4+30\times\frac{15}{16}-16\times(\frac{15}{16})^{2}=4+\frac{450}{16}-\frac{225}{16}=4+\frac{450 - 225}{16}=4+\frac{225}{16}=\frac{64 + 225}{16}=\frac{289}{16}=18.0625\approx18$ feet.

Step4: For Rock B with $g(t)=5 + 20t-16t^{2}$, $a=-16$, $b = 20$, $c = 5$

Calculate the time of maximum height: $t_B=-\frac{b}{2a}=-\frac{20}{2\times(-16)}=\frac{20}{32}=\frac{5}{8}$ seconds.

Step5: Calculate the maximum height of Rock B

$g(\frac{5}{8})=5+20\times\frac{5}{8}-16\times(\frac{5}{8})^{2}=5+\frac{100}{8}-\frac{16\times25}{64}=5+\frac{100}{8}-\frac{25}{4}=5+\frac{100 - 50}{8}=5+\frac{50}{8}=\frac{40+50}{8}=\frac{90}{8}=11.25\approx11$ (the given answer of 16 is incorrect).

Step6: Compare the times to maximum height

Since $\frac{5}{8}=\frac{10}{16}<\frac{15}{16}$, Rock B reaches its maximum height first.

Answer:

a. Rock A: about 18 feet; Rock B: about 11 feet b. Rock B