Answer:
Part a) The function in vertex form is [tex]y=-16(t-2.5)^{2}+106[/tex]
Part b) The maximum height of the arrow is 106 feet
Part c) It takes about 5 seconds for the arrow to hit the ground
Step-by-step explanation:
we have
[tex]y=-16t^{2}+80t+6[/tex]
where
y ----> is the height (in feet) of an arrow
t ----> is the time in seconds after it is shot from a bow
This is the equation of a vertical parabola open downward
The vertex is a maximum
Part a) Part b) Write the function in vertex form. Find the maximum height of the arrow
we have
[tex]y=-16t^{2}+80t+6[/tex]
Factor -16
[tex]y=-16(t^{2}-5t)+6[/tex]
Complete the squares
[tex]y=-16(t^{2}-5t+6.25)+6+100[/tex]
[tex]y=-16(t^{2}-5t+6.25)+106[/tex]
Rewrite as perfect squares
[tex]y=-16(t-2.5)^{2}+106[/tex] ----> equation in vertex form
The vertex is the point (2.5,106)
The maximum height of the arrow represent the y-coordinate of the vertex of the function
therefore
The maximum height of the arrow is 106 feet
Part c) How long does it take the arrow to hit the ground?
we know that
When the arrow hit the ground the value of y is equal to zero
so
For y=0
[tex]0=-16(t-2.5)^{2}+106[/tex]
[tex]16(t-2.5)^{2}=106[/tex]
[tex](t-2.5)^{2}=\frac{106}{16}[/tex]
take square root both sides
[tex](t-2.5)=(+/-)\frac{\sqrt{106}}{4}[/tex]
[tex]t=2.5(+/-)\frac{\sqrt{106}}{4}[/tex]
[tex]t=2.5(+)\frac{\sqrt{106}}{4}=5\ sec[/tex]
[tex]t=2.5(-)\frac{\sqrt{106}}{4}=-0.07\ sec[/tex]
therefore
It takes about 5 seconds for the arrow to hit the ground
