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?

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

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