An astronaut on the moon drops a rock straight downward from a height of 0.95m. If the acceleration due to gravity on the moon is 1.62m/s, what is the speed of the rock just before it lands?
Just give me the answer??
That's not how this works. Where do you have a problem?
Btw, the acceleration should be negative, or else will just keep flying off the moon and
acceleration would be -1.62m/s^2 , not m/s
v = √(2as)
How does this work? well, recall that
s = 1/2 at^2 so, t = √(2s/a)
But at the same time, v = at = a√(2s/a) = √(2as)
So just plug in your numbers, and take the time to learn a few elementary but very useful formulas.
To find the speed of the rock just before it lands, we can use the equations of motion. The equation that relates distance, acceleration, and final velocity is:
v² = u² + 2as
Where:
- v is the final velocity (speed of the rock just before it lands)
- u is the initial velocity (which is 0 since the rock is dropped from rest)
- a is the acceleration due to gravity (given as 1.62 m/s²)
- s is the distance traveled (0.95m in this case, as the rock is dropped from a height of 0.95m)
Substituting the known values into the equation, we get:
v² = 0² + 2(1.62)(0.95)
v² = 0 + 3.09
v² = 3.09
Taking the square root of both sides to solve for v, we get:
v = √3.09
v ≈ 1.76 m/s
Therefore, the speed of the rock just before it lands is approximately 1.76 m/s.