Q. 34 - Precalculus Enhanced with Graphing Utilities

Question

Physics A ball is thrown vertically upward with an initial

velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is $s\left(t\right)=96t-16t^2$.

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

(b) For what time t is the ball more than 128 feet above the ground?

Step-by-Step Solution

Answer

(a) The ball hits the ground after 6 seconds.

(b) The solution set is $\left\{t|2<t<4\right\}$.

Part (a) Step 1. Given Information

The given equation of the distance of the ball from the ground is $s\left(t\right)=96t-16t^2$

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

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

$\begin{array}{rcl}s\left(t\right)& =& 96t-16t^2\\ & =& 16t(6-t)\end{array}$

• Equate both the factors with 0.

$\begin{array}{rcl}16t& =& 0\\ t& =& 0\\ 6-t& =& 0\\ t& =& 6\end{array}$

So, the ball hits the ground after 6 seconds.

Part (b) Step 1. Given Information.

• The given equation is $s\left(t\right)=96t-16t^2$.
• The minimum height of the ball is 128 feet.

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

• The minimum height of the ball is 128 feet. So, $96t-16t^2>128 \mathrm{or} -16t^2+96t-128>0$.
• Find the factors of the function. 

$\begin{array}{rcl}-16t^2+96t-128& =& 0\\ -16(t-2)(t-4)& =& 0\end{array}$

• Equate the factors to 0. 

$\begin{array}{rcl}t-2& =& 0\\ t& =& 2\\ t-4& =& 0\\ t& =& 4\end{array}$

• So, the 0- intercepts are $(2,0),(4,0)$.
• The value of $f\left(0\right)=-128$.
• So, the y- intercept is $(0,-128)$.
• The vertex of the parabola is at $x=\frac{-b}{2a}=\frac{-96}{-32}=3$.
• The value of $f\left(3\right)=16$.
• So, the vertex of the parabola is at $(3,16)$.

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 $2<t<4$.
• So, the solution set is $\left\{t|2<t<4\right\}$