Answer:
The maximum height of the baseball is 9 feet
Step-by-step explanation:
we have
[tex]p(t)=\frac{1}{2}gt^2+v_0t+p_0[/tex]
where
p(t) ----> baseball position above the ground in feet
t ----> the time in seconds
v_0 ----> is the initial velocity in ft/sec
p_0 ---> initial position above the ground
we have
[tex]g=-32\frac{ft}{sec^2} \\\\v_0=16\frac{ft}{sec}\\\\p_0=5\ ft[/tex]
substitute the given values
[tex]p(t)=\frac{1}{2}(-32)t^2+16t+5[/tex]
[tex]p(t)=-16t^2+16t+5[/tex]
This is the equation of a vertical parabola open downward
The vertex represent a maximum
Convert the quadratic equation in vertex form
[tex]p(t)=-16t^2+16t+5[/tex]
Factor -16 leading coefficient
[tex]p(t)=-16(t^2-t)+5[/tex]
Complete the square
[tex]p(t)=-16(t^2-t+\frac{1}{4})+5+4[/tex]
[tex]p(t)=-16(t^2-t+\frac{1}{4})+9[/tex]
Rewrite as perfect squares
[tex]p(t)=-16(t-\frac{1}{2})^2+9[/tex]
The vertex is the point (0.5,9)
The maximum height of the baseball above the ground is the y-coordinate of the vertex
therefore
The maximum height of the baseball is 9 feet
