Answer:
The pack falls approximately 271.44 meters.
Step-by-step explanation:
To solve this problem, we can use the equations of motion under constant acceleration. In this case, the acceleration is due to gravity, which is approximately -9.81 m/s² (negative because it's directed downward). We'll assume that air resistance is negligible.
The equations we'll use are:
Final velocity squared = Initial velocity squared + 2 * acceleration * distance
Distance = Initial velocity * time + 0.5 * acceleration * time^2
Given:
Initial velocity (u) = 0 m/s (the pack is released from rest)
Final velocity (v) = -73.0 m/s (negative because it's directed downward)
Acceleration (a) = -9.81 m/s²
We want to find the distance (s) the pack falls.
Let's use the second equation to solve for distance (s):
v^2 = u^2 + 2as
Substitute the values:
(-73.0 m/s)^2 = (0 m/s)^2 + 2 * (-9.81 m/s²) * s
Simplify and solve for s:
5329 m²/s² = -19.62 m/s² * s
s = 5329 m²/s² / (-19.62 m/s²)
s = -271.44 m
Since distance can't be negative in this context, the negative sign indicates the direction (downward). The pack falls approximately 271.44 meters.