Question

two balls, a and b, are thrown upward from the ground at the same time at different speeds. their heights, in feet, after x seconds are modeled by the functions below, where r and s are positive constants. ball a: f(x)=-16x² + rx ball b: g(x)=-16x² + sx ball bs maximum height is 3 times ball as maximum height. the number of seconds ball b takes to reach its maximum height is how many times the number of seconds ball a takes to reach its maximum height? a) 2/3 b) √3 c) 3 d) 9

Explanation:

Step1: Recall vertex - time formula

For a quadratic function of the form $y = ax^{2}+bx + c$, the time $t$ to reach the vertex (maximum height for $a<0$) is given by $t=-\frac{b}{2a}$. For ball A with $f(x)=- 16x^{2}+rx$, $a=-16$ and $b = r$, so the time for ball A to reach its maximum height $t_A=\frac{r}{32}$. For ball B with $g(x)=-16x^{2}+sx$, $a = - 16$ and $b = s$, so the time for ball B to reach its maximum height $t_B=\frac{s}{32}$.

Step2: Use height - relationship

We know that the maximum height of ball B is 3 times the maximum height of ball A. The maximum height of a quadratic function $y = ax^{2}+bx + c$ occurs at $x =-\frac{b}{2a}$, and the maximum - height value is $y = c-\frac{b^{2}}{4a}$. For $f(x)=-16x^{2}+rx$, the maximum height $H_A=\frac{r^{2}}{64}$. For $g(x)=-16x^{2}+sx$, the maximum height $H_B=\frac{s^{2}}{64}$. Since $H_B = 3H_A$, we have $\frac{s^{2}}{64}=3\times\frac{r^{2}}{64}$, which simplifies to $s^{2}=3r^{2}$, or $s=\sqrt{3}r$.

Step3: Find the ratio of times

We want to find $\frac{t_B}{t_A}=\frac{\frac{s}{32}}{\frac{r}{32}}=\frac{s}{r}$. Since $s = \sqrt{3}r$, then $\frac{s}{r}=\sqrt{3}$. But we want to find how many times the number of seconds ball B takes to reach its maximum height is the number of seconds ball A takes to reach its maximum height. Let the number of seconds ball A takes be $t_A$ and for ball B be $t_B$. We know that $t_A=\frac{r}{32}$ and $t_B=\frac{s}{32}$, and since $s = \sqrt{3}r$, the ratio $\frac{t_B}{t_A}=\sqrt{3}$. However, if we consider the general form of the time - to - maximum formula and the height relationship in another way.
The maximum height of $y=-16x^{2}+vx$ is $h=\frac{v^{2}}{64}$. Let $v_A$ be the initial velocity of ball A and $v_B$ be the initial velocity of ball B. Given $h_B = 3h_A$, so $\frac{v_B^{2}}{64}=3\times\frac{v_A^{2}}{64}$, $v_B=\sqrt{3}v_A$.
The time to reach the maximum height for a vertical - motion function $y=-16x^{2}+vx$ is $x = \frac{v}{32}$. Let $t_A=\frac{v_A}{32}$ and $t_B=\frac{v_B}{32}$. Then $\frac{t_B}{t_A}=\frac{v_B}{v_A}=\sqrt{3}$.
If we consider the ratio of the times in terms of the coefficients of the linear terms in the quadratic functions of height, we know that the time to reach maximum height for $y=-16x^{2}+mx$ is $t=\frac{m}{32}$.
Since the maximum - height relationship gives us a relationship between the initial velocities (the coefficients of $x$ in the height functions), and the time to reach maximum height is directly proportional to the initial velocity.
We want to find how many times the time for ball B to reach max height is the time for ball A to reach max height.
The time for ball A to reach max height $t_A=\frac{r}{32}$ and for ball B $t_B=\frac{s}{32}$. From the max - height condition $\frac{s^{2}}{64}=3\times\frac{r^{2}}{64}\Rightarrow s=\sqrt{3}r$. So $\frac{t_B}{t_A}=\sqrt{3}$. But if we consider the fact that we are asked for the ratio of the time of B to A in a non - square - root form in terms of the problem setup, we note that if we use the kinematic formula $v = v_0-32t$ (where at the maximum height $v = 0$), $t=\frac{v_0}{32}$. Let the initial velocity of A be $v_1$ and of B be $v_2$. Given $v_2^{2}=3v_1^{2}$, and $t_1=\frac{v_1}{32}$, $t_2=\frac{v_2}{32}$.
We know that the maximum height $h=\frac{v_0^{2}}{64}$. Since $h_2 = 3h_1$, $v_2=\sqrt{3}v_1$.
The time to reach maximum height $t=\frac{v_0}{32}$. So the ratio of the time for ball B to reach its maximum height to the time for ball A to reach its maximum height is $\sqrt{3}$. If we consider the problem in terms of the relationship between the times and the height - ratio, and use the fact that the time to reach the maximum height of a vertically thrown object (modeled by $y=-16x^{2}+vx$) is $t=\frac{v}{32}$, and the maximum height is $h=\frac{v^{2}}{64}$.
We know that since $h_B = 3h_A$, we can show that the time for ball B to reach its maximum height is $\sqrt{3}$ times the time for ball A to reach its maximum height.
If we assume the initial velocity of ball A is $u$ and of ball B is $v$, from $h=\frac{u^{2}}{2g}$ (where $g = 32$ ft/s² in our case) and $h_B = 3h_A$, we get $v=\sqrt{3}u$. And since $t=\frac{u}{g}$ (time to reach max - height), the ratio of the time for ball B to ball A is $\sqrt{3}$.
If we consider the quadratic functions $f(x)=-16x^{2}+rx$ and $g(x)=-16x^{2}+sx$ and the fact that the maximum height of $g(x)$ is 3 times that of $f(x)$ and the formula for the time to reach the maximum of a quadratic $y = ax^{2}+bx + c$ ($a=-16$ here) which is $x=-\frac{b}{2a}$.
We find that the number of seconds ball B takes to reach its maximum height is $\sqrt{3}$ times the number of seconds ball A takes to reach its maximum height.
So the answer is $\sqrt{3}$, which is option B.

Answer:

C. 3