Question

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.

Answer 1

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:

`x=(-b)/(2a)`

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:

`h=-16t^(2) +80t+25`

`f(t)=-16t^(2) +80t+25`

`y=-16t^(2) +80t+25`

Where,

`a=-16\\b=80\\c=25`

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

`t=(-b)/(2a)=(-80)/(2(-16))=(-80)/(-32)=2.5`

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

`y=-16(2.5)^(2) +80(2.5)+25`

`y=-16*6.25 +200+25`

`y=-100 +200+25=125`

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!