Question

current objective
solve applications modeled by quadratic equations that may require the quadratic
formula

question
a ball is thrown vertically in the air with a velocity of 90 ft/s. use the projectile formula
$h=-16t^{2}+v_{0}t$ to determine at what time(s), in seconds, the ball is at a height of 120 ft.
round your answer(s) to the nearest tenth of a second.

provide your answer below:

Explanation:

Step1: Substitute values into the formula

Given \(h = 120\), \(v_0=90\), substitute into \(h=-16t^{2}+v_0t\). So \(120=-16t^{2}+90t\). Rearrange to \(16t^{2}-90t + 120=0\). Divide by 2: \(8t^{2}-45t + 60=0\).

Step2: Apply the quadratic formula

For \(ax^{2}+bx + c = 0\) (\(a = 8\), \(b=-45\), \(c = 60\)), \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
\(t=\frac{45\pm\sqrt{(-45)^{2}-4\times8\times60}}{2\times8}=\frac{45\pm\sqrt{2025 - 1920}}{16}=\frac{45\pm\sqrt{105}}{16}\).
\(\sqrt{105}\approx10.25\).
\(t_1=\frac{45 + 10.25}{16}=\frac{55.25}{16}\approx3.45\).
\(t_2=\frac{45-10.25}{16}=\frac{34.75}{16}\approx2.17\).

Answer:

\(2.2\), \(3.5\)