Q. 33 - Precalculus Enhanced with Graphing Utilities

Question

A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is $s\left(t\right)=8t-16t^2$.

(a) At what time t will the ball strike the ground?

(b) For what time t is the ball more than 96 feet above the ground?

Step-by-Step Solution

Answer

(a) The ball hits the ground after 10 seconds.

(b) the solution set is $\left\{x:2<x<3\right\}$.

Part (a) Step 1. Given Information

The given equation of the distance of the ball from the ground is $s\left(t\right)=80t-16t^2$.

Part (b) Step 2. Find t when distance is 0

• Substitute 0 for $s\left(t\right)$ into the given function and then factorize.

$\begin{array}{rcl}0& =& 80t-16t^2\\ 8t(10-t)& =& 0\end{array}$

• Equate both the factors with 0.

$\begin{array}{rcl}8t& =& 0\\ t& =& 0\\ 10-t& =& 0\\ t& =& 10\end{array}$

So, the ball hits the ground after 10 seconds.

Part (b) Step 1. Given Information.

• The given equation is $s\left(t\right)=80t-16t^2$.
• The minimum height of the ball is 96 feet.

Part (b) Step 2. Create a function and find intercepts

• The minimum height of the ball is 96 feet.. So, $80t-16t^2>96 or 80t-16t^2-96>0$.
• Find the factors of the function. 

$\begin{array}{rcl}80t-16t^2-96& =& 0\\ -16(t-2)(t-3)& =& 0\\ -16(t-2)(t-3)& =& 0\end{array}$

• Equate the factors to 0. 

$$\begin{gathered} t-2=0 \\ t=2 \\ t-3=0 \\ t=3 \end{gathered}$$

• So, the 0- intercepts are $(2,0) \mathrm{and} (3,0)$.
• The value of $f\left(0\right)=-96$.
• So, the y- intercept is $(0,-96)$.
• The vertex of the parabola is at $x=\frac{-b}{2a}=\frac{-80}{-32}=\frac{5}{2}$.
• The value of $f\left(\frac{5}{2}\right)=4$.
• So, the vertex of the parabola is at $\left(\frac{5}{2},4\right)$.

Part (b) Step 3. Plot the graph

• Use the intercepts and the vertex to plot the graph of the function.

• From the graph, the curve is above the horizontal axis when $2<x<3$.
• So, the solution set is $\left\{x:2<x<3\right\}$