Answer:
Part 1) The maximum height of the ball is 77.20 feet
Part 2) The maximum height was reached at 2.32 seconds
Part 3) The height of the ball when it was shot was 1.75 feet
Part 4) The ball hit the ground at 4.67 seconds
Part 5) The graph in the attached figure
Step-by-step explanation:
Let
t ----> the time in seconds
h(t) ----> the height of the ball in feet at any time
we have
[tex]h(t)=-14t^2+65t+1.75[/tex]
This is a vertical parabola open downward (the leading coefficient is negative)
The vertex represent a maximum
Part 1) Find the maximum height attained by the ball
we know that
The maximum height correspond to the y-coordinate of the vertex
so
Convert the quadratic equation in vertex form
[tex]h(t)=-14t^2+65t+1.75[/tex]
Factor -14
[tex]h(t)=-14(t^2-\frac{65}{14}t)+1.75[/tex]
Complete the square
[tex]h(t)=-14(t^2-\frac{65}{14}t+\frac{4,225}{784})+1.75+\frac{4,225}{56}[/tex]
[tex]h(t)=-14(t^2-\frac{65}{14}t+\frac{4,225}{784})+\frac{4,323}{56}[/tex]
Rewrite as perfect squares
[tex]h(t)=-14(t-\frac{65}{28})^2+\frac{4,323}{56}[/tex]
the vertex is the point [tex](\frac{65}{28},\frac{4,323}{56})[/tex]
Convert to decimal number
[tex](2.32,77.20)[/tex]
therefore
The maximum height of the ball is 77.20 feet
Part 2) What time was the maximum height achieved?.
we know that
The time when the maximum height was reached is equal to the x-coordinate of the vertex
The vertex is the point [tex](2.32,77.20)[/tex]
therefore
The maximum height was reached at 2.32 seconds
Part 3) How high was the ball off of the ground when it was shot?
Find out the y-intercept
The y-intercept is the height of the ball when it was shot (time equal zero)
so
For t=0
[tex]h(t)=-14(0)^2+65(0)+1.75=1.75\ ft[/tex]
therefore
The height of the ball when it was shot was 1.75 feet
Part 4) How far did the ball travel before it hit the ground?
we know that
When the ball hit the ground, the height is equal to zero
so
For h(t)=0
[tex]-14t^2+65t+1.75=0[/tex]
solve the quadratic equation
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]-14t^2+65t+1.75=0[/tex]
so
[tex]a=-14\\b=65\\c=1.75[/tex]
substitute in the formula
[tex]t=\frac{-65\pm\sqrt{65^{2}-4(-14)(1.75)}} {2(-14)}[/tex]
[tex]t=\frac{-65\pm\sqrt{4,323}} {-28}[/tex]
[tex]t=\frac{-65+\sqrt{4,323}} {-28}=-0.03[/tex]
[tex]t=\frac{-65-\sqrt{4,323}} {-28}=4.67[/tex]
therefore
The ball hit the ground at 4.67 seconds
Part 5) Graph this quadratic
Plot the given points to graph the quadratic equation
we have
open downward
vertex (2.32,77.20) ---> is a maximum
y-intercept (0,1.75)
x-intercepts (-0.03,0) and (4.67,0)
The graph in the attached figure
