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An object drops 22 m from a cliff. If it started at rest and is going 12 m/s12m/s just before it hits the ground, what is its acceleration?

A. 4 m/s²
B. 6 m/s²
C. 8 m/s²
D. 10 m/s²

User Rich Ehmer
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1 Answer

3 votes

Final answer:

To find the acceleration of an object dropped from a cliff, one can use the kinematic equation v^2 = u^2 + 2as. However, when the known values are applied to this equation, the result does not match the available multiple-choice answers, hinting at an issue with the provided numbers or the underlying assumptions.

Step-by-step explanation:

The question pertains to the physics concept of acceleration due to gravity, specifically using one of the kinematic equations to solve for acceleration when an object is dropped from a cliff. To find the acceleration, we can use the kinematic equation v^2 = u^2 + 2as, where

  • v is the final velocity (12 m/s)
  • u is the initial velocity (0 m/s, since it starts from rest)
  • a is the acceleration
  • s is the distance traveled (22 m)

Plugging in the given values and solving for a, we get:

12^2 = 0 + 2*a*22

144 = 44a

a = 144 / 44 = 3.27 m/s²

However, this value doesn't match with any of the provided options A to D. The correct value near to earth's surface should be around 9.8 m/s² signifying a potential error in the given values or calculations.

User Taaam
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