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Zackariaalikhango
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A 23kg rock falls off the edge of a planet that has a surface gravitational field of magnitude 0.167 N/Kg. Assume that the planet has no atmosphere. The speed of the rock just before it strikes the ground is 2.34 m/s.
How tall is the cliff?
Write your answer using three significant figures.
____m

Answer :

Onyebuchinnajigo

The height of the cliff on the given planet is 16.39 m.

The given parameters:

  • Mass of the rock, m = 23 kg
  • Surface gravitational field, g = 0.167 kg N/kg
  • Speed of the rock before it strikes the ground, v = 2.34 m/s

The height of the cliff is determined by applying the principle of conservation of mechanical energy as follows;

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

[tex]h = \frac{v^2}{2g} \\\\h = \frac{(2.34)^2}{2\times 0.167} \\\\h = 16.39 \ m[/tex]

Thus, the height of the cliff on the given planet is 16.39 m.

Learn more about conservation of mechanical energy here: https://brainly.com/question/6852965

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