Chapter 5.4, Problem 46E

a.

To determine

To calculate: The heights that the snowboarder and the kangaroo jump.

Answer to Problem 46E

The kangaroo jumps $2.62\text{ feet}$ and the snowboarder jumps $5.86\text{ feet}$ .

Explanation of Solution

Given information:

Hang time t (in seconds) is given by the function,

  $t=0.5\sqrt{h}$

where h is the height (in feet) of the jump.

Hang time of the kangaroo = $0.81\text{ s}$ .

Hang time of the snowboarder = $1.21\text{ s}$ .

Calculation:

Height that the kangaroo jumps is calculated as:

  $\begin{array}{l}t=0.5\sqrt{h}[\text{Given]}\\ 0.81=0.5\sqrt{h}[\text{Substitute 0}\text{.81 for }t]\\ \sqrt{h}=1.62[\text{Divide both sides by 0}\text{.5]}\\ h^{1/2}=1.62[\text{Express in exponent form]}\\ {\left(h^{1/2}\right)}^2={1.62}^2[\text{Raise both sides to the second power]}\\ h=2.62[\text{Simplify]}\end{array}$

Height that the snowboarder jumps is calculated as:

  $\begin{array}{l}t=0.5\sqrt{h}[\text{Given]}\\ \text{ 1}\text{.2}1=0.5\sqrt{h}[\text{Substitute 1}\text{.21 for }t]\\ \sqrt{h}=2.42[\text{Divide both sides by 0}\text{.5]}\\ h^{1/2}=2.42[\text{Express in exponent form]}\\ {\left(h^{1/2}\right)}^2={2.42}^2[\text{Raise both sides to the second power]}\\ h=5.86[\text{Simplify]}\end{array}$

b.

To determine

To calculate: The heights that the snowboarder and the kangaroo jump when hang time is doubled.

Answer to Problem 46E

The kangaroo jumps $\text{10}\text{.50 feet}$ and the snowboarder jumps $\text{23}\text{.43 feet}$ when hang times of both are doubled .

Explanation of Solution

Given information:

Hang time t (in seconds) is given by the function,

  $t=0.5\sqrt{h}$

where h is the height (in feet) of the jump.

Hang time of the kangaroo = $0.81\text{ s}$ .

Hang time of the snowboarder = $1.21\text{ s}$ .

Calculation:

New hang time of the of the kangaroo = $0.81\times 2\text{ s = 1}\text{.62 s}$

New hang time of the of the kangaroo = $1.21\times 2\text{ s = 2}\text{.42 s}$

Height that the kangaroo jumps is calculated as:

  $\begin{array}{l}t=0.5\sqrt{h}[\text{Given]}\\ \text{ 1}\text{.62}=0.5\sqrt{h}[\text{Substitute 1}\text{.62 for }t]\\ \sqrt{h}=3.24[\text{Divide both sides by 0}\text{.5]}\\ h^{1/2}=3.24[\text{Express in exponent form]}\\ {\left(h^{1/2}\right)}^2={3.24}^2[\text{Raise both sides to the second power]}\\ h=10.50[\text{Simplify]}\end{array}$

Height that the snowboarder jumps is calculated as:

  $\begin{array}{l}t=0.5\sqrt{h}[\text{Given]}\\ \text{ 2}\text{.42}=0.5\sqrt{h}[\text{Substitute 2}\text{.42 for }t]\\ \sqrt{h}=4.84[\text{Divide both sides by 0}\text{.5]}\\ h^{1/2}=4.84[\text{Express in exponent form]}\\ {\left(h^{1/2}\right)}^2={4.84}^2[\text{Raise both sides to the second power]}\\ h=23.43[\text{Simplify]}\end{array}$

c.

To determine

To explain: If the height of the jump doubles when hang time doubles.

Answer to Problem 46E

The height of the jump becomes four times for both, the kangaroo and the snowboarder when their hang times are doubled.

Explanation of Solution

Given information:

Hang time t (in seconds) is given by the function,

  $t=0.5\sqrt{h}$

where h is the height (in feet) of the jump.

Hang time of the kangaroo = $0.81\text{ s}$ .

Hang time of the snowboarder = $1.21\text{ s}$ .

Calculation:

Height that the kangaroo jumps in the given hang time = $\text{2}\text{.62 feet}$

Height that the kangaroo jumps when hang time is doubled = $\text{10}\text{.50 feet}$

Number of times the height increases when hang time is doubled is calculated as:

  $\frac{\text{10}\text{.50 feet}}{2.62\text{ feet}}=4$

Height that the snowboarder jumps in the given hang time = $\text{5}\text{.86 feet}$

Height that the kangaroo jumps when hang time is doubled = $\text{23}\text{.43 feet}$

Number of times the height increases when hang time is doubled is calculated as:

  $\frac{\text{23}\text{.43 feet}}{\text{5}\text{.86 feet}}=4$