Chapter 4, Problem 20E

A 30.0-kg child starting from rest slides down a water slide with a vertical height of 10.0 m. What is the child’s speed (a) halfway down the slide’s vertical distance and (b) three-fourths of the way down? (Neglect friction.)

To determine

The speed of the child while coming down the halfway down the slide in vertical distance.

Answer to Problem 20E

The speed of the child while coming down the halfway down the slide in vertical distance is $69.3\text{ m}\cdot {\text{s}}^{-1}$.

Explanation of Solution

Given info: The speed of the child having mass $30.0\text{ kg}$ is coming down the halfway down the slide from rest with vertical height of $10.0\text{ m}$.

Explanation:

Formula to calculate the speed of an object while coming from a height is,

$v=\sqrt{2g\Delta h}$

Here,

$\Delta h$ is the change in height

$g$ is the acceleration due to gravity

$v$ is the speed of an object

Calculate the height at the halfway down the water slide.

$h_2=\frac{1}{2}{h}_1$

Here,

$h_1,h_2$ are the heights

Substitute $10.0\text{ m}$ for $h_1$.

$\begin{array}{c}{h}_2=\frac{1}{2}\left(10.0\text{ m}\right)\\ =5\text{ m}\end{array}$

Calculate the change in height.

$\Delta h=h_1-h_2$

Substitute $10.0\text{ m}$ for $h_1$ and $5.0\text{ m}$ for $h_2$.

$\begin{array}{c}\Delta h=10.0\text{ m}-5.0\text{ m}\\ =\text{5 m}\end{array}$

Calculate the speed of the child using the formula of the speed of an object while coming from a height.

$v=\sqrt{2g\Delta h}$

Substitute $9.8\text{ m}\cdot {\text{s}}^{-2}$ for $g$ and $\text{5 m}$ for $\Delta h$.

$\begin{array}{c}v=\sqrt{2\left(9.8\text{ m}\cdot {\text{s}}^{-2}\right)\left(\text{5 m}\right)}\\ =69.3\text{ m}\cdot {\text{s}}^{-1}\end{array}$

Thus, the speed of the child while coming down the halfway down the slide in vertical distance is $69.3\text{ m}\cdot {\text{s}}^{-1}$.

Conclusion:

Therefore, thus, the speed of the child while coming down the halfway down the slide in vertical distance is $69.3\text{ m}\cdot {\text{s}}^{-1}$.

To determine

The speed of the child while coming down the three-fourth down the slide in vertical distance.

Answer to Problem 20E

The speed of the child while coming down the three-fourth down the slide in vertical distance is $34.6\text{ m}\cdot {\text{s}}^{-1}$.

Explanation of Solution

Given info: The speed of the child having mass $30.0\text{ kg}$ is coming down from the rest from the slide. He comes down the three-fourth down the slide with vertical height of $10.0\text{ m}$.

Explanation:

Formula to calculate the speed of an object while coming from a height is,

$v=\sqrt{2g\Delta h}$

Here,

$\Delta h$ is the change in height.

$g$ is the acceleration due to gravity.

$v$ is the speed of an object.

Calculate the height at the three fourth down the water slide.

$h_2=\frac{3}{4}{h}_1$

Here,

$h_1,h_2$ are the heights

Substitute $10.0\text{ m}$ for $h_1$.

$\begin{array}{c}{h}_2=\frac{3}{4}\left(10.0\text{ m}\right)\\ =7.5\text{ m}\end{array}$

Calculate the change in height.

$\Delta h=h_1-h_2$

Substitute $10.0\text{ m}$ for $h_1$ and $7.5\text{ m}$ for $h_2$.

$\begin{array}{c}\Delta h=10.0\text{ m}-7.5\text{ m}\\ =2.\text{5 m}\end{array}$

Calculate the speed of the child using the formula of the speed of an object while coming from a height.

$v=\sqrt{2g\Delta h}$

Substitute $9.8\text{ m}\cdot {\text{s}}^{-2}$ for $g$ and $2.\text{5 m}$ for $\Delta h$.

$\begin{array}{c}v=\sqrt{2\left(9.8\text{ m}\cdot {\text{s}}^{-2}\right)\left(2.\text{5 m}\right)}\\ =34.6\text{ m}\cdot {\text{s}}^{-1}\end{array}$

Thus, the speed of the child while coming down the halfway down the slide in vertical distance is $34.6\text{ m}\cdot {\text{s}}^{-1}$.

Conclusion:

Therefore, thus, the speed of the child while coming down the halfway down the slide in vertical distance is $34.6\text{ m}\cdot {\text{s}}^{-1}$.