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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

Respuesta :

The answer to this question is 5 seconds.

The given function is

[tex] b(t) = 80t-16t^2+3.5 [/tex]

Rearranging it we get

[tex] b(t) = -16t^2+80t+3.5 [/tex]

Now when the ball hits the ground

Height b(t) becomes 0.

So we have

[tex] -16t^2+80t+3.5=0 [/tex]

Comparing with

[tex] ax^2+bx+c [/tex]

We get

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

Substituting in the quadratic formula we get

[tex] t=\frac{-b+\sqrt{b^2-4ac}}{2a} or t=\frac{-b-\sqrt{b^2-4ac}}{2a}

[/tex][tex] t=\frac{-80+\sqrt{80^2-4(-16)(3.5)}}{2(-16)} =-0.04 [/tex]

OR

[tex] t=\frac{-80-\sqrt{80^2-4(-16)(3.5)}}{2(-16)} =5.04 [/tex]

But the time cannot be negative

So t = 5.04 seconds

Rounding to the nearest second we get

t = 5 seconds

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