An astronaut on Neptune drops a rock straight downward from a height of 0.95 m. If the acceleration of gravity on Neptune is 11.2 m/s2, what is the speed of the rock just before it lands? (m/s)
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A dropped object attains a velocity of
V = sqrt(2 a X)
when falling a distance X.
Use the Neptune gravity acceleration value for a.
4.61
To find the speed of the rock just before it lands on the surface of Neptune, we can use the laws of motion. The equation that relates distance, acceleration, initial velocity, and final velocity is:
v^2 = u^2 + 2as
Where:
v = final velocity (speed of the rock just before it lands)
u = initial velocity (which is 0 because the rock is dropped)
a = acceleration due to gravity on Neptune (11.2 m/s^2)
s = distance traveled (0.95 m)
Since the initial velocity (u) is 0, we can simplify the equation to:
v^2 = 2as
Substituting the given values:
v^2 = 2 * 11.2 m/s^2 * 0.95 m
v^2 = 20.96 m^2/s^2
Taking the square root of both sides:
v = √(20.96 m^2/s^2)
v ≈ 4.58 m/s
Therefore, the speed of the rock just before it lands on Neptune is approximately 4.58 m/s.