Respuesta :
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
