Question

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the acceleration due to gravity near Earth's surface, the acceleration due to gravity near the surface of the planet is approximately
A) the same
B) twice as great
C) one-half as great
D) four times as great

Answer 1

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

$\Delta x=v_0t+\frac{gt^2}{2}$

In this case, the sphere starts from rest, so $v_0=0$. Replacing the given values and solving for g':

$g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}$

The acceleration due to gravity near Earth's surface is $g=9.8\frac{m}{s^2}$. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.