Answer:
part 1) 0.78 seconds
part 2) 1.74 seconds
Step-by-step explanation:
step 1
At about what time did the ball reach the maximum?
Let
h ----> the height of a ball in feet
t ---> the time in seconds
we have
[tex]h(t)=-16t^{2}+25t+5[/tex]
This is a vertical parabola open downward (the leading coefficient is negative)
The vertex represent a maximum
so
The x-coordinate of the vertex represent the time when the ball reach the maximum
Find the vertex
Convert the equation in vertex form
Factor -16
[tex]h(t)=-16(t^{2}-\frac{25}{16}t)+5[/tex]
Complete the square
[tex]h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}[/tex]
[tex]h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}[/tex]
Rewrite as perfect squares
[tex]h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}[/tex]
The vertex is the point [tex](\frac{25}{32},\frac{945}{64})[/tex]
therefore
The time when the ball reach the maximum is 25/32 sec or 0.78 sec
step 2
At about what time did the ball reach the minimum?
we know that
The ball reach the minimum when the the ball reach the ground (h=0)
For h=0
[tex]0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}[/tex]
[tex]16(t-\frac{25}{32})^{2}=\frac{945}{64}[/tex]
[tex](t-\frac{25}{32})^{2}=\frac{945}{1,024}[/tex]
square root both sides
[tex](t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}[/tex]
[tex]t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}[/tex]
the positive value is
[tex]t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec[/tex]
