Chapter 3, Problem 25P

To determine

The distance jumped by a person can jump on the moon as compared to Earth.

Answer to Problem 25P

6 times farther on moon as compared to earth.

Explanation of Solution

Given:

The initial velocity on moon and earth is same and angle of projection for both is same acceleration due to gravity on moon is one-sixth

Formula Used :

Horizontal Range :

  $R=u^2\frac{\sin 2\theta }{g}$

Calculation :

  $\text{Initial velocity on moon = }{u}_e$

  $\text{Angle on moon = }{\theta }_m$

  $\text{Angle on earth = }{\theta }_e$

Substituting the values

  $R_e=u_e{}^2\frac{\sin 2{\theta }_e}{g_e}$

  $R_m=u_m{}^2\frac{\sin 2{\theta }_m}{g_m}$

  ${\theta }_m={\theta }_e\text{=}\theta \text{ and }{u}_e^2=u_m^2=u^2$

  $g_m=\frac{1}{6}{g}_e$

  $\frac{R_e}{R_m}=u^2\frac{\sin 2\theta }{g_e}\times \frac{g_m}{u^2\sin 2\theta }$

  $\begin{array}{l}\frac{R_e}{R_m}=\frac{g_m}{g_e}\\ \frac{R_e}{R_m}=\frac{g_e}{6g_e}\end{array}$

  $\frac{R_e}{R_m}=\frac{1}{6}$

  $R_m=6R_e$

Conclusion :

Thus, a person can jump 6 times farther on moon as compared to earth.