Question

jen served the final point before winning her tennis match! she tossed the ball up into the air from a height of 5 feet with an initial velocity of 22 feet per second. after the ball started to come back down, jen hit the ball with her racket at a height of 6 feet. which equation can you use to find how many seconds the ball was in the air before jen hit it? if an object travels upward with an initial velocity of v feet per second from s feet above the ground, the objects height in feet, h, after t seconds can be modeled by the formula h = -16t²+vt + s. 6 = -16t²+22t + 5 5 = -16t²+22t + 6 to the nearest tenth of a second, how long was the ball in the air before jen hit it? seconds

Explanation:

Step1: Identify the quadratic - equation

The height of the ball is modeled by the quadratic equation \(h=-16t^{2}+22t + 5\). When the ball hits the racket at a height of \(h = 6\), we substitute \(h = 6\) into the equation, getting \(6=-16t^{2}+22t + 5\).

Step2: Rearrange the equation

Rearrange \(6=-16t^{2}+22t + 5\) to the standard quadratic - form \(ax^{2}+bx + c = 0\). So, \(16t^{2}-22t+1 = 0\). Here, \(a = 16\), \(b=-22\), and \(c = 1\).

Step3: Use the quadratic formula

The quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 16\), \(b=-22\), and \(c = 1\) into the formula:
\[

$$\begin{align*} t&=\frac{22\pm\sqrt{(-22)^{2}-4\times16\times1}}{2\times16}\\ &=\frac{22\pm\sqrt{484 - 64}}{32}\\ &=\frac{22\pm\sqrt{420}}{32}\\ &=\frac{22\pm2\sqrt{105}}{32}\\ &=\frac{11\pm\sqrt{105}}{16} \end{align*}$$

\]

Step4: Calculate the values of \(t\)

\(t_1=\frac{11+\sqrt{105}}{16}\approx\frac{11 + 10.25}{16}=\frac{21.25}{16}\approx1.3\) and \(t_2=\frac{11-\sqrt{105}}{16}\approx\frac{11 - 10.25}{16}=\frac{0.75}{16}\approx0.05\). We take the non - zero value (since \(t = 0\) is the starting time), so \(t\approx1.3\) seconds.

Answer:

\(1.3\) seconds