Question

An Astronaut on the Moon drops a rock straight downward from a height of 1.25 meters. If the acceleration of gravity on the Moon is 1.62 meters per second squared, what is the speed of the rock just before it lands?

Answer 1

Answer:

Since the astronaut drops the rock, the initial velocity of the rock is 0 m/s

We are given:

initial velocity (u) = 0 m/s

final velocity (v) = v m/s

acceleration (a) = 1.62 m/s/s

height (h) = 1.25 m

Solving for v:

From the third equation of motion:

v²-u² = 2ah

replacing the variables

v² - (0)² =2 (1.62)(1.25)

v² = 1.62 * 2.5

v² = 4 (approx)

v = √4

v = 2 m/s

The speed of the rock just before it lands is 2 m/s