Answer:
51.4 m/s
Explanation:
vf^2 = 2(23.6m/s^2)(56.0m)
vf = sqrt[2(23.6m/s^2)(56.0m)]
vf = 51.4 m/s
The speed of an object when it hits the ground from a given height on a given planet is required.
The speed of the boulder before it strikes the ground is 51.41 m/s.
m = Mass of boulder = 650 kg
s = Displacement = 56 m
u = Initial velocity = 0
v = Final velocity
Surface gravitational field = [tex]23.6\ \text{N/kg}=23.6\dfrac{\dfrac{\text{kg m}}{\text{s}^2}}{\text{kg}}=23.6\ \text{m/s}^2[/tex]
This means surface gravitational field is acceleration due to gravity on the planet.
g = [tex]23.6\ \text{m/s}^2[/tex]
From the kinematic equations of motion we have
[tex]v^2-u^2=2gs\\\Rightarrow v=\sqrt{2gs+u^2}\\\Rightarrow v=\sqrt{2\times 23.6\times 56+0}\\\Rightarrow v=51.41\ \text{m/s}[/tex]
The speed of the boulder before it strikes the ground is 51.41 m/s.
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