Question

A rock hits the ground w/a speed of 8m/s and comes to a stop embedded in the ground 0.055seconds later. How deep is the hole made by the rock?
(Not a free-fall problem).

Answer 1

Answer 2

To find the depth of the hole made by the rock, we can use the equations of motion for constant acceleration. The rock experiences constant deceleration as it comes to a stop.

Let's denote the initial velocity of the rock as u (8 m/s, but negative since it is decelerating), the final velocity as v (0 m/s), the time taken as t (0.055 seconds), and the depth of the hole as d.

The equations of motion are:

1. v = u + at

2. d = ut + (1/2)at^2

We need to find the depth (d). The final velocity (v) is 0 m/s, the initial velocity (u) is -8 m/s, and the time taken (t) is 0.055 seconds.

Step 1: Find acceleration (a) using the equation v = u + at 0 = -8 + a * 0.055 a = 8 / 0.055 a ≈ 145.45 m/s^2

Step 2: Find the depth (d) using the equation d = ut + (1/2)at^2 d = (-8) * 0.055 + (1/2) * 145.45 * (0.055)^2 d = -0.44 + 0.1125 d ≈ -0.3275 meters

Since depth cannot be negative, we take the magnitude of the depth, which is approximately 0.3275 meters.

So, the hole made by the rock is approximately 0.3275 meters deep.