Solved Problem on One-dimensional Motion

On the Moon, a stone is released, at rest from a height of 20 meters. It falls, under the action of the lunar acceleration due to gravity, until reaching the ground with a speed v. Find how high the stone should be dropped on Earth, so that it hits the ground at the same speed v. The acceleration due to gravity on Earth g_E = 9.8 m/s², acceleration due to gravity on the Moon g_M = 1.6 m/s².

Problem data:

Height of fall: H_M = 20 m;
Initial speed of the stone: v₀ = 0;
Acceleration due to gravity on Earth: g_E = 9.8 m/s²;
Acceleration due to gravity on the Moon: g_M = 1.6 m/s².

Problem diagram:

We choose a frame of reference oriented upwards with origin on the ground. In both cases, the accelerations due to gravity point to the ground, and their signals will be negative.

Figure 1

For the Moon, we have the initial position of the stone S_0M = 20 m and the final position S_M = 0, for the Earth, the initial position will be S_0E = H_E, and the final position will be S_E = 0.

Solution

To find the final speed with which the stone hits the ground, since we do not have the time interval of fall, we use the equation of velocity as a function of displacement

\[ \begin{gather} \bbox[#99CCFF,10px] {v^2=v_0^2+2a\Delta S} \end{gather} \]

For the Moon:

\[ \begin{gather} v_{\small M}^2=v_0^2+2g_{\small M}\Delta S_{\small M}\\[5pt] v_{\small M}^2=v_0^2+2g_{\small M}(S_{\small M}-S_{0\small M})\\[5pt] v_{\small M}^2=0-2\times\left(1.6\;\mathrm{\small{\frac{m}{s^2}}}\right)(0-20\;\mathrm m)\\[5pt] v_{\small M}=\sqrt{64\;\mathrm{\small{\frac{m^2}{s^2}}}\;}\\[5pt] v_{\small M}=8\;\mathrm{m/s} \end{gather} \]

We want the speed at which the stone reaches the ground, on Earth, to be the same speed as the Moon, v = v_M = v_E.

For the Earth:

\[ \begin{gather} v_{\small M}^2=v_0^2+2g_{\small E}\Delta S_{\small E}\\[5pt] v_{\small M}^2=v_0^2+2g_{\small E}(S_{\small E}-H_{\small E})\\[5pt] S_{\small E}-H_{\small E}=\frac{v_{\small M}^2-v_0^2}{2g_{\small E}}\\[5pt] -H_{\small E}=\frac{v_{\small M}^2-v_0^2}{2g_{\small E}}-S_{\small E}\\[5pt] -H_{\small E}=\frac{\left(8\;\mathrm{\frac{m}{s}}\right)^2-0^2}{2\times\left(-9,8\;\mathrm{\frac{m}{s^2}}\right)}-0\\[5pt] -H_{\small E}=\frac{64\;\mathrm{\frac{m^\cancel 2}{\cancel{s^2}}}}{-19,6\;\mathrm{\frac{\cancel m}{\cancel{s^2}}}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {H_{\small E}\approx 3.3\;\mathrm m} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .