Chapter 9.2, Problem 30PPE

Solution

(a)

To calculate: The initial height of the geyser.

The required height is $25875\text{ft}$ .

Given Information:

The given velocity is $69\text{ft}/\text{s}$ .

Time taken is 6.25 hr.

Formula Used:

Velocity of an object is

  $\text{velocity}=\frac{\text{distance}}{\text{time}}$

Calculation:

It is given that velocity is $69\text{ft}/\text{s}$ and the time is 6.25 hr.

Convert hour into second.

  $\begin{array}{c}1hr=60s\\ 6.25hr=6.25\times 60s\\ =375s\end{array}$

Now, find the height. Use the formula $\text{velocity}=\frac{\text{distance}}{\text{time}}$ .

  $\begin{array}{c}\text{height}=69\times 375\\ =25875\text{ft}\end{array}$

Hence, the required height is $25875\text{ft}$ .

(b)

To find: The expression of the function that represent the height $h$ in $t$ seconds after it start erupting.

The required expression is $h=69t+25875$ .

Given Information:

The given velocity is $69\text{ft}/\text{s}$ .

Formula Used:

Velocity of an object is

  $\text{velocity}=\frac{\text{distance}}{\text{time}}$

Explanation:

It is given that the velocity of the geyser is $69\text{ft}/\text{s}$ .

It is known that $\text{distande}=\text{velocity}\times \text{time}$ .

Thus, for height $h$ and time $t$ and the velocity $69\text{ft}/\text{s}$ . the expression will be,

  $h=69t$ .

Since, the water travel from the height of $25875\text{ft}$ to 0 from the ground.

Hence, the required expression is $h=69t+25875$ .