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 a 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 at a 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. 5 = -16t² + 22t + 6 6 = -16t² + 22t + 5 to the nearest tenth of a second, how long was the ball in the air before jen hit it? seconds

Explanation:

Step1: Identify variables

Given: s=5, v=22, h=6. Formula: \( h = -16t^2 + vt + s \).

Step2: Substitute into formula

\( 6 = -16t^2 + 22t + 5 \).

Step3: Rearrange to quadratic equation

\( 16t^2 - 22t + 1 = 0 \).

Step4: Solve using quadratic formula

\( t = \frac{22 \pm \sqrt{(-22)^2 - 4(16)(1)}}{2(16)} = \frac{22 \pm \sqrt{420}}{32} \).

Step5: Select valid solution

Positive roots: ~0.05s (up) and ~1.3s (down). Choose ~1.3s.

Answer:

6 = -16t² + 22t + 5
1.3