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?
V = sqrt(2 a X) when the initial velocity is zero.
a is the acceleration rate
X is the distance fallen
Solve for V
To find the speed of the rock just before it lands, we need to use the equations of motion. The equation that relates the speed (v), initial velocity (u), acceleration (a), and distance (s) is:
v^2 = u^2 + 2as
In this case, the initial velocity (u) is 0 m/s since the rock is dropped, and the acceleration (a) is 1.62 m/s^2 (gravity on the Moon). The distance (s) is the height from which the rock was dropped, which is 1.25 meters.
Using the equation, we can substitute the known values:
v^2 = 0^2 + 2(1.62)(1.25)
Simplifying:
v^2 = 0 + 4.05
v^2 = 4.05
To find v, we take the square root of both sides:
v = √4.05
v ≈ 2.012 m/s
Therefore, the speed of the rock just before it lands is approximately 2.012 meters per second.
To find the speed of the rock just before it lands, you can use the equations of motion. Specifically, you can use the equation that relates the final velocity (v), initial velocity (u), acceleration (a), and displacement (s).
In this case, the rock is dropped vertically downward, so the initial velocity (u) is zero. The acceleration due to gravity on the Moon is given as 1.62 meters per second squared (a). The rock falls from a height of 1.25 meters (s).
The equation you can use is:
v^2 = u^2 + 2as
Plugging in the known values:
v^2 = 0 + 2 * 1.62 * 1.25
Simplifying:
v^2 = 4.05
To find v, you take the square root of both sides:
v = √4.05
Calculating:
v ≈ 2.01 meters per second
Therefore, the speed of the rock just before it lands is approximately 2.01 meters per second.