Question

1. we know that the gravitational acceleration on the moon is one - sixth that of the earths gravitational acceleration. how long does it take a 2.50 kg rock to fall 15.0 m on the moon?

Explanation:

Step1: Determine the gravitational acceleration on the moon

The gravitational acceleration on Earth \(g_{earth}=9.8\ m/s^{2}\).
Since \(g_{moon}=\frac{1}{6}g_{earth}\), then \(g_{moon}=\frac{1}{6}\times9.8\ m/s^{2}\approx1.63\ m/s^{2}\).

Step2: Use the kinematic equation \(h = v_{0}t+\frac{1}{2}at^{2}\)

The rock is in free - fall, so the initial velocity \(v_{0} = 0\ m/s\), and the displacement \(h = 15.0\ m\), and \(a = g_{moon}\).
Substituting into the equation \(h=\frac{1}{2}g_{moon}t^{2}\), we can solve for \(t\).
First, rewrite the equation for \(t\): \(t=\sqrt{\frac{2h}{g_{moon}}}\).
Substitute \(h = 15.0\ m\) and \(g_{moon}\approx1.63\ m/s^{2}\) into the formula:
\(t=\sqrt{\frac{2\times15.0}{1.63}}\)
\(t=\sqrt{\frac{30}{1.63}}\approx\sqrt{18.40}\approx 4.29\ s\)

Answer:

It takes approximately \(4.29\ s\) for the rock to fall.