A rock falls off a cliff and hits the ground after three seconds. The rock's velocity is 29.4 m/s when it hits the ground.

What is its acceleration of the rock in the downward direction?

A.0.31 m/s2
B. 3 m/s2
C. 9.8 m/s2
D. 88.2 m/s2

1 answer

To solve the problem, we can use the information given about the rock's fall:

  1. The rock falls for a total time of 3 seconds.
  2. Its final velocity just before hitting the ground is 29.4 m/s.

Using the equation of motion relating final velocity (\(v\)), initial velocity (\(u\)), acceleration (\(a\)), and time (\(t\)):

\[ v = u + at \]

Assuming the rock starts from rest, the initial velocity \(u = 0\). Thus, the equation simplifies to:

\[ v = at \]

Substituting the values we have:

\[ 29.4 , \text{m/s} = a \cdot 3 , \text{s} \]

To solve for \(a\), divide both sides by 3 seconds:

\[ a = \frac{29.4 , \text{m/s}}{3 , \text{s}} = 9.8 , \text{m/s}^2 \]

Thus, the acceleration of the rock in the downward direction is:

C. 9.8 m/s²