Chapter 3, Problem 20P

Solution

To Determine:The factor by which a person can jump farther on the Moon as compared to that on the Earth.

Solution:A person on the Moon reaches 6 times more distance than on the Earth.

Explanation:

Given:

The initial speed of the person on the Moon $v_{i,M}$ is the same as the initial speed on Earth $v_{i,E}$ and the launch angleof the person on the Moon ${\theta }_{i,M}$ is the same as the launch angle on Earth ${\theta }_{i,E}$ . The gravity on the Moon $g_M$ is one-sixth on Earth $g_E$ .

Formula used:

The horizontal range R is defined as:

  $R=\frac{v_i{}^2\sin 2\theta }{g}$

Where:

• R is the horizontal range.
• $v_i$ is the initial speed of the person.
• $\theta$ is the launch angle.
• g is the acceleration due to gravity.

Calculation:

For the person on Earth, the above equation can be written as:

  $R_E=\frac{v_{iE}{}^2\sin 2{\theta }_E}{g_E}$

Similarly, for the person on Moon:

  $R_M=\frac{v_{iM}{}^2\sin 2{\theta }_M}{g_M}$

But:

  $v_{i,E}=v_{i,M}=v_i$

And

  ${\theta }_E={\theta }_M=\theta$

  $R_E=\frac{v_i{}^2\sin 2\theta }{g_E}$

And

  $R_M=\frac{v_i{}^2\sin 2\theta }{g_M}$

Take the ratio

  $\frac{R_M}{R_E}=\frac{g_E}{g_M}$

As $g_M=\frac{1}{6}{g}_E$

  $\Rightarrow R_M=6R_E$

Conclusion:

Thus, a person on the Moon reaches 6 times more distance than on Earth.