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)=-16x^{2}+rx ball b: g(x)=-16x^{2}+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. $\frac{2}{3}$ b. $sqrt{3}$ c. 3 d. 9

Explanation:

Step1: Recall the formula for the vertex of a parabola

The height - time function of a projectile is a quadratic function of the form \(y = ax^{2}+bx + c\), and the \(x\) - coordinate of the vertex (which gives the time to reach the maximum height) is \(t=-\frac{b}{2a}\). For ball A, \(f(x)=-16x^{2}+rx\), where \(a=-16\) and \(b = r\), so the time for ball A to reach its maximum height \(t_A=-\frac{r}{2\times(-16)}=\frac{r}{32}\). For ball B, \(g(x)=-16x^{2}+sx\), where \(a = - 16\) and \(b = s\), so the time for ball B to reach its maximum height \(t_B=-\frac{s}{2\times(-16)}=\frac{s}{32}\).

Step2: Use the maximum - height relationship

The maximum height of a quadratic function \(y = ax^{2}+bx + c\) is \(y=\frac{4ac - b^{2}}{4a}\). The maximum height of ball A, \(H_A=\frac{0 - r^{2}}{4\times(-16)}=\frac{r^{2}}{64}\), and the maximum height of ball B, \(H_B=\frac{0 - s^{2}}{4\times(-16)}=\frac{s^{2}}{64}\). Given that \(H_B = 3H_A\), so \(\frac{s^{2}}{64}=3\times\frac{r^{2}}{64}\), which simplifies to \(s^{2}=3r^{2}\), and since \(s\) and \(r\) are positive, \(s = r\sqrt{3}\).

Step3: Find the ratio of the times

We want to find \(\frac{t_B}{t_A}=\frac{\frac{s}{32}}{\frac{r}{32}}=\frac{s}{r}\). Substituting \(s = r\sqrt{3}\) into the ratio, we get \(\frac{t_B}{t_A}=\sqrt{3}\).

Answer:

B. \(\sqrt{3}\)