Chapter 15.3, Problem 9CFU

a.

To determine

Find the distance travelled by the ball in first $6\sec onds$ .

Answer to Problem 9CFU

  $576units$

Explanation of Solution

Given information:

An object in free fall accelerates at $32feet/{\sec }^2$ , the velocity $t\sec onds$ after dropped is $32feet/\sec$ . Neglect the air resistance. The ball is dropped from the top of Sears Tower.

  

Calculation:

Consider the given velocity,

  $v\left(t\right)=32t$

Now, calculate the distance travelled by the ball in first $6\sec onds$ ,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{6}{\int }}32tdt}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}32t_i}\Delta t\\ \Delta t=\frac{6-0}{n}=\frac{6}{n},\\ t_1=a+i\Delta t=0+\frac{6i}{n}\\ so,\\ {\displaystyle \underset{0}{\overset{6}{\int }}32tdt}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}32\left(\frac{6i}{n}\right)}\left(\frac{6}{n}\right)\\ =1152\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i}{n^2}\right)}\\ =1152\underset{n\to \infty }{\lim }\frac{1}{n^2}{\displaystyle \sum _{i=1}^{n}i}\\ =1152\underset{n\to \infty }{\lim }\frac{1}{n^2}\frac{n\left(n+1\right)}{2}\\ =1152\underset{n\to \infty }{\lim }\frac{1}{n^2}\frac{n^2\left(1+\frac{1}{n}\right)}{2}\\ =576\end{array}$

Hence the distance travelled by the ball in first $6\sec onds$ is, $576units$ .

b.

To determine

Verify that the ball hit the ground within $10\sec onds$ fall from tower.

Answer to Problem 9CFU

The distance travelled by the ball in first $10\sec onds$ is, $1600units$ .it is greater than the Tower height, the ball must the hit the ground in less than $10\sec onds$ .

Explanation of Solution

Given information:

An object in free fall accelerates at $32feet/{\sec }^2$ , the velocity $t\sec onds$ after dropped is $32feet/\sec$ . Neglect the air resistance. The ball is dropped from the top of Sears Tower.

  

Calculation:

Consider the given velocity,

  $v\left(t\right)=32t$

Now, calculate the distance travelled by the ball in first $10\sec onds$ ,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{10}{\int }}32tdt}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}32t_i}\Delta t\\ \Delta t=\frac{10-0}{n}=\frac{10}{n},\\ t_1=a+i\Delta t=0+\frac{10i}{n}\\ so,\\ {\displaystyle \underset{0}{\overset{10}{\int }}32tdt}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}32\left(\frac{10i}{n}\right)}\left(\frac{10}{n}\right)\\ =3200\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i}{n^2}\right)}\\ =3200\underset{n\to \infty }{\lim }\frac{1}{n^2}{\displaystyle \sum _{i=1}^{n}i}\\ =3200\underset{n\to \infty }{\lim }\frac{1}{n^2}\left(1+2+\mathrm{...}+n\right)\\ =3200\underset{n\to \infty }{\lim }\frac{1}{n^2}\frac{n^2\left(1+\frac{1}{n}\right)}{2}\\ =1600\end{array}$

Hence the distance travelled by the ball in first $10\sec onds$ is, $1600units$ .it is greater than the Tower height, the ball must the hit the ground in less than $10\sec onds$ .