Q46. - Algebra 2

Question

An object is fired straight up from the top of a 200-foot tower at a velocity of 80 feet per second. The height h(t) of the object t seconds after firing is given by $h\left(t\right)=-16t^2+80t+200$. Find the maximum height reached by the object and the time that the height is reached.

Step-by-Step Solution

Answer

The maximum height reached by the object is 300 feet and the time at which maximum height is reached is 2.5 seconds.

Step 1. Given Information.

The given function is $h\left(t\right)=-16t^2+80t+200$

Step 2. Use the concept.

Consider the function $f\left(x\right)=a{x}^2+bx+c,a\ne 0$, the x-coordinate of vertex is $\frac{-b}{2a}$.

 

The Graph of $f\left(x\right)=a{x}^2+bx+c,a\ne 0$

 

• opens up and has a minimum value when $a>0$, and
• opens down and has a maximum value when $a<0$.

Step 3. Solution.

In the function $h\left(t\right)=-16t^2+80t+200$, we have $a=-16,b=80,c=200$

Here, $a=-16<0$

So, the graph opens down and has a maximum value.

The maximum value of the function is the y-coordinate of the vertex and is the maximum height reached by the object.

The x-coordinate of the vertex is

 $\begin{array}{l}\frac{-b}{2a}\\ =\frac{-\left(80\right)}{2(-16)}\mathrm{..........}\left[a=-16,b=80\right]\\ =2.5\end{array}$

Find the y-coordinate of the vertex by evaluating the function for $x=2.5$.

$h(2.5)=-16{\left(2.5\right)}^2+80\left(2.5\right)+200=300$

So, the maximum height reached by the object is 300 at 2.5 seconds.