Answer:
The bird will take 0.88 seconds to reach a height of 40 feet in the air for the first time
Step-by-step explanation:
we have
[tex]h=-16t^2+55t+4[/tex]
This is a vertical parabola open downward
The vertex represent a maximum
where
h is the height of the bird
t is the amount of time, in seconds, the bird had been in the air
For h=40 ft
substitute in the quadratic equation
[tex]40=-16t^2+55t+4[/tex]
[tex]-16t^2+55t+4-40=0[/tex]
[tex]-16t^2+55t-36=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-16t^2+55t-36=0[/tex]
so
[tex]a=-16\\b=55\\c=-36[/tex]
substitute in the formula
[tex]x=\frac{-55\pm\sqrt{55^{2}-4(-16)(-36)}} {2(-16)}[/tex]
[tex]x=\frac{-55\pm\sqrt{721}} {-32}[/tex]
[tex]x=\frac{-55+\sqrt{721}} {-32}=0.88\ sec[/tex]
[tex]x=\frac{-55-\sqrt{721}} {-32}=2.56\ sec[/tex]
therefore
The bird will take 0.88 seconds to reach a height of 40 feet in the air for the first time
