a rock hits the ground at a speed of 10m/s and leaves a hole 25cm deep. what is the magnitude of the (assumed) uniform deceleration of the rock?

To find the magnitude of the uniform deceleration of the rock, we can use the kinematic equation:

vf^2 = vi^2 + 2ad

Where:
- vf is the final velocity (0 m/s since the rock comes to rest)
- vi is the initial velocity (10 m/s)
- a is the uniform deceleration we are trying to find
- d is the displacement (25 cm = 0.25 m)

Let's plug in the given values:

0 = (10)^2 + 2a(0.25)

Simplifying the equation:

0 = 100 + 0.5a

Rearranging the equation to solve for "a":

0.5a = -100

a = -100 / 0.5

a = -200 m/s^2

Hence, the magnitude of the assumed uniform deceleration of the rock is 200 m/s^2.

To find the magnitude of the deceleration of the rock, we can use the equation of motion:

\(v^2 = u^2 + 2as\)

where:
- \(v\) is the final velocity (which is 0 as the rock comes to a stop),
- \(u\) is the initial velocity (10 m/s),
- \(a\) is the deceleration (what we are trying to find), and
- \(s\) is the displacement (25 cm or 0.25 m, since it's given in centimeters).

Let's rearrange the equation to solve for the deceleration:

\(0 = (10 \, \text{m/s})^2 + 2a \cdot (0.25 \, \text{m})\)

\(0 = 100 \, \text{m/s}^2 + 0.5a\)

Rearranging again:

\(0.5a = -100 \, \text{m/s}^2\)

\(a = \frac{-100 \, \text{m/s}^2}{0.5}\)

\(a = -200 \, \text{m/s}^2\)

Therefore, the magnitude of the uniform deceleration of the rock is 200 m/s². Note that the negative sign indicates that the deceleration is in the opposite direction of the initial velocity.

Vf^2=Vi^2 + 2ad solve for a.

2,000 m/s^2