Chapter 1.1, Problem 60PS

Solution

a.

To find: The functions for the path of the ball hit on earth and on moon.

The functions for the path of the ball on Earth is $y_E=-\frac{2}{625}{x}^2+x$ and on the moon is $y_E=-\frac{32}{10,000}{x}^2+x$ .

Given information:

The path of a ball hit an angle ${45}^{\circ }$ with a speed of 100 ft per second is modeled by $y=-\frac{g}{10,000}{x}^2+x$ , where $x$ is the horizontal distance in feet and $y$ is the corresponding height and $g$ is the acceleration due to gravity.

Concept used:

Substitute the value of acceleration due to gravity to find the functions on different grounds.

Calculation:

Substitute $g=32$ to find the equation of the ball on Earth.

  $\begin{array}{c}{y}_E=-\frac{32}{10,000}{x}^2+x\\ =-\frac{2}{625}{x}^2+x\end{array}$

Now, substitute $g=16$ to find the function of the ball on moon.

  $\begin{array}{c}{y}_m=-\frac{5.3}{10,000}{x}^2+x\\ =-0.00053x^2+x\end{array}$

Conclusion:

The functions for the path of the ball on Earth is $y_E=-\frac{2}{625}{x}^2+x$ and on the moon is $y_m=-0.00053x^2+x$ .

b.

To find: Graph the functions obtained in part (a) and, then find the distances travelled by the ball on Earth and on moon.

Graphs of the functions can be given as shown below.

  

The maximum distance travelled by the ball on Earth is 312.5 feet and on moon is 1886.792 feet.

Given information:

The path of a ball hit an angle ${45}^{\circ }$ with a speed of 100 ft per second is modeled by $y=-\frac{g}{10,000}{x}^2+x$ , where $x$ is the horizontal distance in feet and $y$ is the corresponding height and $g$ is the acceleration due to gravity.

Concept used:

The maximum distance of the ball whose path is given by a quadratic function is the distance between the x-intercept of the graph.

Calculation:

Use graphing calculator to graph the functions for the path of the ball on Earth and on moon.

  

The red curve shows the path $y_E=-\frac{2}{625}{x}^2+x$ and the blue curve shows the path $y_m=-0.00053x^2+x$ .

Observe that x-intercept of the function $y_E=-\frac{2}{625}{x}^2+x$ is $\left(312.5,0\right)$ and the x-intercept of the function $y_m=-0.00053x^2+x$ is $\left(1886.792,0\right)$ .

So, the maximum distance travelled by the ball on Earth is 312.5 feet and on moon is 1886.792 feet.

Conclusion:

The maximum distance travelled by the ball on Earth is 312.5 feet and on moon is 1886.792 feet.

c.

To interpret: The distance travelled by the ball on Earth and on the moon in terms of the ratio of the distances traveled and how the distances and values of $g$ are related.

The distance traveled by the ball on Earth is less than the distance traveled on the moon as the acceleration due to gravity on Earth is higher than that of the moon.

Given information:

The path of a ball hit an angle ${45}^{\circ }$ with a speed of 100 ft per second is modeled by $y=-\frac{g}{10,000}{x}^2+x$ , where $x$ is the horizontal distance in feet and $y$ is the corresponding height and $g$ is the acceleration due to gravity.

Concept used:

The maximum distance of the ball whose path is given by a quadratic function is the distance between the x-intercept of the graph.

Calculation:

Find the ratio of the distance traveled by the ball on Earth to the distance traveled on moon.

  $\begin{array}{c}\frac{\text{Distance traveled on Earth}}{\text{Distance traveled on the moon}}=\frac{312.5}{1886.792}\\ =0.1656\end{array}$

Observe that $0.1656<1$ . So, the distance traveled by the ball on Earth far less than the distance traveled on the moon.

The acceleration due to gravity on Earth is higher than the acceleration due to gravity on the moon. This shows that higher the acceleration due gravity less is the distance traveled.

Conclusion:

The distance traveled by the ball on Earth is less than the distance traveled on the moon as the acceleration due to gravity on Earth is higher than that of the moon.