Question

The CHS baseball team wasn’t on the field and the batter popped the ball up.The equation b(t)= 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

The given function is

$b(t) = 80t-16t^2+3.5$

Rearranging it we get

$b(t) = -16t^2+80t+3.5$

Now when the ball hits the ground

Height b(t) becomes 0.

So we have

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

Comparing with

$ax^2+bx+c$

We get

a=-16, b=80 & c = 3.5

Substituting in the quadratic formula we get

$t=\frac{-b+\sqrt{b^2-4ac}}{2a} or t=\frac{-b-\sqrt{b^2-4ac}}{2a}$$t=\frac{-80+\sqrt{80^2-4(-16)(3.5)}}{2(-16)} =-0.04$

OR

$t=\frac{-80-\sqrt{80^2-4(-16)(3.5)}}{2(-16)} =5.04$

But the time cannot be negative

So t = 5.04 seconds

Rounding to the nearest second we get

t = 5 seconds

Answer 2

The answer to this question is 5 seconds.