April shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high. The pathway of the arrow can be represented by the equation h=-16t squared+80t+25, where h is the highest and t is the time in seconds. What is the maximum height of the arrow.

Respuesta :

Hello!

The answer is: The maximum height of the arrow is 125 feet.

Why?

To solve this problem, we need to know the motion of the arrow is described by a quadratic equation (parabola). Also, we are given a parabola with a negative coefficient. So, to solve this problem, we need to remember the following:

- To know if the parabola is opening upward or downward, we need to look the sign of the coefficient of the quadratic term.

- We need to remember that the highest or lowest point of a parabola is given by the y-coordinate of the vertex.

We can calculate the vertex using the following formula:

[tex]x=\frac{-b}{2a}[/tex]

Then, to find the y-coordinate of the vertex, we need to substitute the x value into the parabola equation.

So, we are given the pathway of the arrow represented by the quadratic equation:

[tex]h=-16t^{2} +80t+25[/tex]

[tex]f(t)=-16t^{2} +80t+25[/tex]

[tex]y=-16t^{2} +80t+25[/tex]

Where,

[tex]a=-16\\b=80\\c=25[/tex]

So, calculating the vertex coordinates in order to find the maximum height of the arrow (highest point), we have:

[tex]t=\frac{-b}{2a}=\frac{-80}{2(-16)}=\frac{-80}{-32}=2.5[/tex]

Then, substituting "t" into the equation of the parabola, we have:

[tex]y=-16(2.5)^{2} +80(2.5)+25[/tex]

[tex]y=-16*6.25 +200+25[/tex]

[tex]y=-100 +200+25=125[/tex]

Therefore, we have that the vertex coordinates are (2.5,125) being the y-coordinate the highest point of the parabola, which is equal to the maximum height of the arrow.

Hence, the maximum height of the arrow is 125 feet.

Have a nice day!

Have a nice day!