Question

on earth, an object is launched horizontally with an initial speed $v$ at a height $h$ and lands at a horizontal distance $x_e$ from its launch point. an astronaut on the surface of the moon horizontally launches an object with the same initial speed $v$ and height $h$, and this object lands at a horizontal distance $x_m$ from its launch point. what is the ratio $\frac{x_e}{x_m}$ in terms of the acceleration of gravity $g_e$ near earths surface and $g_m$ near the moons surface?
(a) $sqrt{\frac{g_m}{g_e}}$
(b) $sqrt{\frac{g_e}{g_m}}$
(c) $\frac{g_m}{g_e}$
(d) $\frac{g_e}{g_m}$

Explanation:

Step1: Analyze vertical - motion

The vertical - motion of the launched object is a free - fall motion. Using the equation $h = v_{0y}t+\frac{1}{2}gt^{2}$, since the object is launched horizontally, $v_{0y} = 0$. So, $h=\frac{1}{2}gt^{2}$, and we can solve for the time of flight $t=\sqrt{\frac{2h}{g}}$.

Step2: Analyze horizontal - motion

The horizontal motion of the launched object is a uniform - motion with constant speed. The horizontal distance $x = v_{0x}t$. Since $v_{0x}=v$ (the initial horizontal speed), on Earth $x_E = v\sqrt{\frac{2h}{g_E}}$, and on the Moon $x_M = v\sqrt{\frac{2h}{g_M}}$.

Step3: Calculate the ratio

We want to find $\frac{x_E}{x_M}$. Substitute the expressions for $x_E$ and $x_M$ into the ratio:
\[

$$\begin{align*} \frac{x_E}{x_M}&=\frac{v\sqrt{\frac{2h}{g_E}}}{v\sqrt{\frac{2h}{g_M}}}\\ &=\sqrt{\frac{g_M}{g_E}} \end{align*}$$

\]

Answer:

A. $\sqrt{\frac{g_M}{g_E}}$