A 15 kg object is released from rest at a height of 800 meters above the ground. The air resistance is proportional to the cube of the speed, with a proportionality constant of k = 0.02kg / (m ^ 2) Determine the speed of the object after 5 seconds and the distance it has fallen by that time.

We are tasked with determining the speed and the distance fallen by a 15 kg object after 5 seconds, given that it is subject to air resistance proportional to the cube of its velocity, and the object was released from rest at a height of 800 meters.

### Step 1: Force Equation

The forces acting on the object are:
1. Gravitational force: \( F_{\text{gravity}} = mg \)
   where \( m = 15 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \) is the gravitational acceleration.
   
   Therefore:
   \[
   F_{\text{gravity}} = 15 \times 9.8 = 147 \, \text{N}
   \]

2. Air resistance: \( F_{\text{drag}} = k v^3 \), where \( k = 0.02 \, \text{kg/m}^2 \) and \( v \) is the velocity of the object.

The net force acting on the object at any time is:
\[
F_{\text{net}} = F_{\text{gravity}} - F_{\text{drag}}
\]
Using Newton's second law \( F = ma \), we have:
\[
m \frac{dv}{dt} = mg - kv^3
\]
which simplifies to:
\[
15 \frac{dv}{dt} = 147 - 0.02 v^3
\]

Dividing both sides by 15, we get:
\[
\frac{dv}{dt} = 9.8 - \frac{0.02}{15} v^3
\]
\[
\frac{dv}{dt} = 9.8 - 0.00133 v^3
\]
This is a nonlinear differential equation that governs the velocity of the object.

### Step 2: Solve the Differential Equation (Numerically)

Since this is not straightforward to solve analytically, we will solve this differential equation numerically to find the velocity \( v \) after 5 seconds and then calculate the distance traveled in that time. Let me compute this.After 5 seconds, the speed of the object is approximately \( 19.40 \, \text{m/s} \), and the distance it has fallen by that time is approximately \( 73.93 \, \text{m} \).

### Summary:
- Speed after 5 seconds: \( 19.40 \, \text{m/s} \)
- Distance fallen after 5 seconds: \( 73.93 \, \text{m} \)

Would you like further details on any part of the calculation?

### Related Questions:
1. How would the speed change if air resistance was proportional to the square of the velocity instead?
2. What would the speed and distance fallen be after 10 seconds?
3. How would the initial height affect the final velocity when the object reaches the ground?
4. What is the terminal velocity for this object under the given air resistance law?
5. How does the value of the proportionality constant \( k \) influence the motion?

### Tip:
For objects in free fall, the drag force plays a significant role in determining their terminal velocity, which can be estimated when drag balances the gravitational force.

Learn how to calculate the speed and distance fallen by a 15 kg object released from 800 meters after 5 seconds with air resistance proportional to the cube of its velocity. The solution involves differential equations and Newton's laws of motion. Suitable for university-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation