Chapter 11.2, Problem 49E

Solution

a)

To draw: The scatter diagram of based on given points in the $xy-$ plane.

The scatter diagram is shown in Figure (1).

Given information:

A ball is pushed off the roof of the building. The velocity (ft/sec) given for different values in the interval $\left[0,1.4\right]$ .

Calculation:

Plot the points $\left(x,y\right)$ in $xy-$ plane. The scatter diagram is shown in Figure (1).

  

Figure (1)

Hence, the required scattered diagram is shown above.

b)

To find: The height of the building using RRAM approach.

The height of the building is approximately 34 ft.

Given information:

A ball is pushed off the roof of the building. The velocity (ft/sec) given for different values in the interval $\left[0,1.4\right]$ .

Formula used:

The area of the rectangle is $base\times height$ .

Right Rectangular Approximation Method (RRAM):

Divide an interval $\left[a,b\right]$ into $n$ subintervals of length $\Delta x=\frac{\left(b-a\right)}{n}$ . The subintervals are $\left[a,a+\Delta x\right]$ , $\left[a+\Delta x,a+2\Delta x\right],$ …, $\left[a+\left(n-1\right)\Delta x,a+n\Delta x=b\right]$ . The area under the curve approximated as the sum of the products defined below:

  ${\displaystyle \sum _{i=1}^{n}f\left(a+i\Delta x\right)\Delta x}$

Calculation:

Divide the interval $\left[0,1.4\right]$

into seven sub-intervals. The velocity of the right endpointsof the subintervals is considered as the height of the rectangle and the base of each rectangle is 0.2. The area of seven rectangles and the total area are computed in the table below:

Subinterval	Base of Rectangle	Height of Rectangle	Area of Rectangle
$\left[0,0.2\right]$	0.2	-5.05	$\left(0.2\right)\left(-5.05\right)=-1.01$
$\left[0.2,0.4\right]$	0.2	$-11.43$	$\left(0.2\right)\left(-11.43\right)=-2.29$
$\left[0.4,0.6\right]$	0.2	$-17.46$	$\left(0.2\right)\left(-17.46\right)=-3.49$
$\left[0.6,0.8\right]$	0.2	$-24.21$	$\left(0.2\right)\left(-24.21\right)=-4.85$
$\left[0.8,1.0\right]$	0.2	$-30.62$	$\left(0.2\right)\left(-30.62\right)=-6.12$
$\left[1.0,1.2\right]$	0.2	$-37.06$	$\left(0.2\right)\left(-37.06\right)=-7.41$
$\left[1.2,1.4\right]$	0.2	$-43.47$	$\left(0.2\right)\left(-43.47\right)=-8.69$
		Total area:	$-33.86$

The absolute value of the total area is 33.86 square feet. The distance traveled by the ball in 1.4 sec is equal to the area above the velocity function. Thus, the distance traveled by the ball in 1.4 sec is 33.86 ft.

Hence, the height of the building is approximately 33.86 ft.