Chapter 11.4, Problem 23E

Solution

a.

To find: The velocity of the ball at $t=1.5\sec$ by using the average velocity on the interval $\left[1,2\right]$ .

The average velocity of the ball at $t=1.5\sec$ is $-50\text{ft}/\sec$ .

Given information: The ball is dropped from the roof of a $30$ -story building. The height of the falling ball in feet above the ground is measured at $0.5$ sec intervals.

Time (sec)	Height (ft)
$0$	$500$
$0.5$	$495$
$1.0$	$485$
$1.5$	$465$
$2.0$	$435$
$2.5$	$400$
$3.0$	$355$
$3.5$	$305$
$4.0$	$245$
$4.5$	$175$
$5.0$	$100$
$5.5$	$15$

Calculation:

From the table, the height of the ball at $1$ sec is $485$ and the height of the ball at $2$ sec is $435$ .

Calculate the average velocity of the ball on the interval $\left[1,2\right]$ .

  $\begin{array}{c}\frac{\Delta s}{\Delta t}=\frac{435-485}{2-1}\\ =-\frac{50}{1}\\ =-50\end{array}$

Thus, the average velocity of the ball at $t=1.5\sec$ is $-50\text{ft}/\sec$ .

b.

To graph: The scatter plot of the given data.

The scatter plot of the given data is shown in figure (1).

Given information: The ball is dropped from the roof of a $30$ -story building. The height of the falling ball in feet above the ground is measured at $0.5$ sec intervals.

Time (sec)	Height (ft)
$0$	$500$
$0.5$	$495$
$1.0$	$485$
$1.5$	$465$
$2.0$	$435$
$2.5$	$400$
$3.0$	$355$
$3.5$	$305$
$4.0$	$245$
$4.5$	$175$
$5.0$	$100$
$5.5$	$15$

Graph:

Plot all the points given in the table on the coordinate plane as shown below.

  

Figure (1)

Thus, the scatter plot of the given data is shown in figure (1).

c.

To find: The quadratic regression model for the given data.

The quadratic regression equation for the given data is $y=-16.08x^2+0.36x+499.77$ .

Given information: The ball is dropped from the roof of a $30$ -story building. The height of the falling ball in feet above the ground is measured at $0.5$ sec intervals.

Time (sec)	Height (ft)
$0$	$500$
$0.5$	$495$
$1.0$	$485$
$1.5$	$465$
$2.0$	$435$
$2.5$	$400$
$3.0$	$355$
$3.5$	$305$
$4.0$	$245$
$4.5$	$175$
$5.0$	$100$
$5.5$	$15$

Calculation:

The general equation of the quadratic regression model is $y=a{x}^2+bx+c$ .

From the table, the value of $a=-16.08$ , $b=0.36$ and $c=499.77$ .

The required quadratic equation is given as:

  $y=-16.08x^2+0.36x+499.77$

Thus, the required quadratic regression equation for the given data is $y=-16.08x^2+0.36x+499.77$ .

d.

To find: The velocity of the ball at $t=1.5\sec$ .

The velocity of the ball at $t=1.5\sec$ is $-47.85\text{ft}/\sec$ .

Given information: The ball is dropped from the roof of a $30$ -story building. The height of the falling ball in feet above the ground is measured at $0.5$ sec intervals.

Time (sec)	Height (ft)
$0$	$500$
$0.5$	$495$
$1.0$	$485$
$1.5$	$465$
$2.0$	$435$
$2.5$	$400$
$3.0$	$355$
$3.5$	$305$
$4.0$	$245$
$4.5$	$175$
$5.0$	$100$
$5.5$	$15$

Calculation:

From part (c), the quadratic equation for the given data is $y=-16.08x^2+0.36x+499.77$ .

Apply the NDER method to find the velocity.

  $\begin{array}{c}\text{NDER }f\left(x\right)=\frac{f\left(1.5+0.001\right)-f\left(1.5-0.001\right)}{0.002}\\ =\frac{f\left(1.501\right)-f\left(1.499\right)}{0.002}\\ =\frac{464.0821-464.1778}{0.002}\\ =-\frac{0.096}{0.002}\end{array}$

Further simplify the above value.

  $\text{NDER }f\left(x\right)=-47.85$

Thus, the velocity of the ball at $t=1.5\sec$ is $-47.85\text{ft}/\sec$ .

e.

To find: The speed of the ball when it hits the ground.

The speed of the ball when it hits the ground is $-179.28\text{ft}/\sec$ .

Given information: The ball is dropped from the roof of a $30$ -story building. The height of the falling ball in feet above the ground is measured at $0.5$ sec intervals.

Time (sec)	Height (ft)
$0$	$500$
$0.5$	$495$
$1.0$	$485$
$1.5$	$465$
$2.0$	$435$
$2.5$	$400$
$3.0$	$355$
$3.5$	$305$
$4.0$	$245$
$4.5$	$175$
$5.0$	$100$
$5.5$	$15$

Calculation:

From part (c), the quadratic equation for the given data is $y=-16.08x^2+0.36x+499.77$ .

Draw the graph of the quadratic regression as shown below.

  

Figure (2)

From the above figure, it can be observed that the ball hits the ground at $t=5.586\sec$ .

Apply the NDER method to find the velocity.

  $\begin{array}{c}\text{NDER }f\left(x\right)=\frac{f\left(5.586+0.001\right)-f\left(5.586-0.001\right)}{0.002}\\ =\frac{f\left(5.587\right)-f\left(5.585\right)}{0.002}\\ =-179.28\end{array}$

Thus, the speed of the ball when it hits the ground is $-179.28\text{ft}/\sec$ .