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
