Answer:
Part A) The maximum height of the basketball is 18 feet
Part B) The basketball is in the air 2.06 seconds
Step-by-step explanation:
Part A) Find the maximum height of the basketball
Let
y is the height of a basketball in feet
t is the time in seconds
we have
[tex]y=-16t^{2} +32t+2[/tex]
This is a vertical parabola open downward (the leading coefficient is negative)
The vertex represent a maximum
Find the vertex
Convert the quadratic equation into vertex form
[tex]y=-16t^{2} +32t+2[/tex]
Factor -16
[tex]y=-16(t^{2} -2t)+2[/tex]
Complete the square
[tex]y=-16(t^{2} -2t+1)+2+16[/tex]
[tex]y=-16(t^{2} -2t+1)+18[/tex]
Rewrite as perfect squares
[tex]y=-16(t-1)^{2}+18[/tex]
The vertex is the point (1,18)
That means
The maximum height is 18 ft when t=1 sec
therefore
The maximum height of the basketball is 18 feet
Part B) How long is the basketball in the air
Find the x-intercepts of the quadratic equation
The x-intercepts are the values of x when the value of y is equal to zero
In this context, the x-intercept is the time in second when the height of the basketball is zero (on the floor)
so
For y=0
[tex]0=-16(t-1)^{2}+18[/tex]
[tex]16(t-1)^{2}=18[/tex]
simplify
[tex](t-1)^{2}=1.125[/tex]
square root both sides
[tex]t-1=\pm\sqrt{1.125}[/tex]
[tex]t=1+\pm\sqrt{1.125}[/tex]
[tex]t=1+\sqrt{1.125}=2.06\ sec[/tex]
[tex]t=1-\sqrt{1.125}=-0.06\ sec[/tex] ---> is not a solution (time cannot be negative)
therefore
The basketball is in the air 2.06 seconds
