Question

The CHS Baseball team was on the field and the batter popped the ball up. The equation b(x)=80t-16t^2+3.5 represents the height of the ball above the ground in feet as a function of time in seconds. How long will the catcher have to get in position to catch the ball before it hits the ground? Round to the nearest second.

Answer 1

Answer:

t = 5 seconds

Step-by-step explanation:

We must solve the equation of movement of the ball as a function of time in seconds.

We must find how many seconds it takes until the ball touches the ground. When the ball touches the ground, the height b (t) = 0

Then we equate the equation to zero and solve for t.

$80t-16t ^ 2 + 3.5 = 0$

Use the quadratic formula.

Where

$a = -16\\\\b = 80\\\\c = 3.5$

$t = \frac{-b \±\sqrt{b^2-4ac}}{2a}\\\\\\t_1 = \frac{-(80)-\sqrt{(80)^2-4(-16)(3.5)}}{2(-16)}\\\\t_1= 5.04\ s\\-------------------------\\t_2 =\frac{-(80)+\sqrt{(80)^2-4(-16)(3.5)}}{2(-16)}\\\\t_2=-0.0434\ s$

We take the positive solution

t = 5 seconds