Chapter 2, Problem 58GP

To determine

The height gained by an object thrown on Moon as a factor of the height gained by same object on Earth thrown with same initial velocity

Answer to Problem 58GP

  $6$ Times

Explanation of Solution

Given:

For Earth:

Initial velocity of object $=v_{eo}=v_o$

Final velocity of the object $=v_{ef}=0\frac{\text{m}}{\text{s}}$

Height gained by object $=h_e$

Displacement of object $=\Delta y_e=h_e$

Acceleration of the object $=a_e=g$

For Moon:

Initial velocity of object $=v_{mo}=v_o$

Final velocity of the object $=v_{mf}=0\frac{\text{m}}{\text{s}}$

Height gained by object $=h_m$

Displacement of object $=\Delta y_m=h_m$

Acceleration of the object $=a_m=\frac{g}{6}$

Formula Used:

Displacement of the object in accelerated motion is given as

  $\Delta y=\frac{v_f^2-v_o^2}{2a}$

Calculation:

Displacement of the object on earth is given as

  $\begin{array}{l}\Delta y_e=\frac{v_{ef}^2-v_{eo}^2}{2a_e}\\ h_e=\frac{0^2-v_o^2}{2g}\\ h_e=\frac{-v_o^2}{2g}\left(\text{1}\right)\end{array}$

Displacement of the object on moon is given as

  $\begin{array}{l}\Delta y_m=\frac{v_{mf}^2-v_{mo}^2}{2a_m}\\ h_m=\frac{0^2-v_o^2}{2\left(\frac{g}{6}\right)}\\ h_m=\frac{-6v_o^2}{2g}\left(\text{2}\right)\end{array}$

Divide equation (2) by equation (1).

  $\begin{array}{l}\frac{h_m}{h_e}=\frac{\left(\frac{-6v_o^2}{2g}\right)}{\left(\frac{-v_o^2}{2g}\right)}\\ \frac{h_m}{h_e}=6\\ h_m=6h_e\end{array}$

Conclusion:

Therefore, height gained is six times the height gained on Earth