Item 10 The height y (in feet) of an arrow t seconds after it is shot from a bow can be modeled by the function y=−16t2+80t + 6. a. Write the function in vertex form. y= b. Find the maximum height of the arrow. The maximum height of the arrow is feet. c. How long does it take the arrow to hit the ground? Round your answer to the nearest second. It takes about seconds for the arrow to hit the ground.

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 feetPart c)  It takes about 5 seconds for the arrow to hit the groundStep-by-step explanation:we have[tex]y=-16t^{2}+80t+6[/tex]wherey ---->  is the height (in feet) of an arrowt ----> is the time in seconds after it is shot from a bowThis is the equation of a vertical parabola open downwardThe vertex is a maximumPart a) Part b) Write the function in vertex form. Find the maximum height of the arrowwe 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 formThe vertex is the point (2.5,106)The maximum height of the arrow represent the y-coordinate of the vertex of the functionthereforeThe maximum height of the arrow is 106 feetPart c) How long does it take the arrow to hit the ground?we know thatWhen the arrow hit the ground the value of y is equal to zerosoFor 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