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 below , where is the height and t is the time in seconds. What is the maximum height of the arrow

Question

h = -16t ^2 + 80t + 25

Answer 1

To find the maximum height of the arrow represented by the equation $h(t)=-16t^2+80t+25$, we can use the formula for the vertex of a parabola, since the equation is in the form of a quadratic equation $h(t)=a{t}^2+bt+c$.

The vertex of a parabola given by $h(t)$ occurs at $t=-\frac{b}{2a}$. In this case:

• $a=-16$
• $b=80$

Substituting these values into the vertex formula:

$$t=-\frac{80}{2\times -16}=-\frac{80}{-32}=\frac{80}{32}=\frac{5}{2}=2.5\text{ seconds}$$

Now that we have the time at which the maximum height occurs, we can substitute $t=2.5$ back into the height equation to find the maximum height $h(2.5)$:

$$h(2.5)=-16(2.5{)}^2+80(2.5)+25$$

Calculating $(2.5{)}^2$:

$$(2.5{)}^2=6.25$$

Now substitute this value into the equation:

$$h(2.5)=-16(6.25)+80(2.5)+25$$ $$=-100+200+25$$ $$=125\text{ feet}$$

The maximum height of the arrow is $\boxed{{\displaystyle 125}}$ feet.