Chapter 3.6, Problem 60PS

Solution

i.

To state: The heights that the basketball player and the kangaroo jumped if "Hang time" is the time you are suspended in the air during a jump.

The heights that the basketball player and the kangaroo jumped are $2.6244\text{m}\text{and}5.0176\text{m}$ respectively.

Given information:

The function is $t=0.5\sqrt{h}$ .

Hang time of the basketball player is $t_b=0.81\sec$ and that of kangaroo is $t_k=1.12\sec$ .

Calculation:

Write the given equation:

  $t=0.5\sqrt{h}$

Divide both sides of the equation by 0.5,

  $\begin{array}{l}\frac{t}{0.5}=\frac{0.5\sqrt{h}}{0.5}\\ \frac{t}{0.5}=\sqrt{h}\end{array}$

Square both sides of the equation,

  $\begin{array}{c}{\left(\frac{t}{0.5}\right)}^2={\left(\sqrt{h}\right)}^2\\ \frac{t^2}{0.25}=h\end{array}$

Now substitute $t_b$ as 0.81 to find the height of the jump by basketball player,

  $\begin{array}{c}{h}_b=\frac{t_b{}^2}{0.25}\\ \frac{{\left(0.81\right)}^2}{0.25}\text{m=}\frac{0.6561}{0.25}\text{m}\\ \text{=0}\text{.6244m}\end{array}$

Now substitute $t_k$ as 1.12 to find the height of the jump by kangaroo,

  $\begin{array}{c}{h}_k=\frac{t_k{}^2}{0.25}\\ \frac{{\left(1.12\right)}^2}{0.25}\text{m=}\frac{1.2544}{0.25}\text{m}\\ \text{=5}\text{.0176m}\end{array}$

Therefore, the jump height of basketball player and that of kangaroo are $2.6244\text{m}\text{and}5.0176\text{m}$ respectively.

ii.

To state: The corresponding heights of each jump after doubling the hang times of the basketball player and the kangaroo.

The new heights that the basketball player and the kangaroo jumped are $\text{10}\text{.4976}\text{m}\text{and}20.0704\text{m}$ respectively.

Given information:

Double the hang times of the basketball player and the kangaroo.

New hang time of the basketball player is:

  $\begin{array}{c}{t}_b=\left(0.81\right)\cdot 2\\ =1.62\sec \end{array}$

New hang time of the kangaroo is:

  $\begin{array}{c}{t}_k=\left(1.12\right)\cdot 2\\ =2.24\sec \end{array}$

Explanation:

Now substitute $t_b$ as 1.62 seconds to find the height of the jump by basketball player,

  $\begin{array}{c}{h}_b=\frac{t_b{}^2}{0.25}\\ \frac{{\left(1.62\right)}^2}{0.25}\text{m=}\frac{2.6244}{0.25}\text{m}\\ \text{=10}\text{.4976m}\end{array}$

Now substitute $t_k$ as 2.24 seconds to find the height of the jump by kangaroo,

  $\begin{array}{c}{h}_k=\frac{t_k{}^2}{0.25}\\ \frac{{\left(2.24\right)}^2}{0.25}\text{m=}\frac{5.0176}{0.25}\text{m}\\ \text{=20}\text{.0704m}\end{array}$

Therefore, the new jump height of basketball player and that of kangaroo are $\text{10}\text{.4976}\text{m}\text{and}20.0704\text{m}$ respectively.

iii.

To state: If the hang time doubles, does the height of the jump double? Explain.

Doubling the hang time results in 4 times of the jump height.

Given information:

The statement says, “Does doubling the hang time double the height of the jump?”

Explanation:

Original heights of the jump are:

Basketball player: $2.6244\text{m}$

Kangaroo: $5.0176\text{m}$

New heights of the jump after doubling the hang time are:

Basketball player: $\text{10}\text{.4976}\text{m}$

Kangaroo: $20.0704\text{m}$

Now check if the height doubles after doubling the hang time:

For the basketball player:

  $\frac{\text{10}\text{.4976}\text{m}}{2.6244m}=4\text{times}$

For the kangaroo:

  $\frac{\text{20}\text{.0704}\text{m}}{5.0176\text{m}}=4\text{times}$

It is clear that doubling the hang time results in 4 times of the jump height.

The relation between the hang time and the jump height is shown below:

  $h=\frac{t^2}{0.25}$

The height of the jump is not directly proportional to the hang time. It is proportional to the square of the hang time.