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Part 1
An object of mass 5 kg is given an initial downward velocity of 50 ​m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is minus20​v, where v is the velocity of the object in​ m/sec. Determine the equation of motion of the object. If the object is initially 400 m above the​ ground, determine when the object will strike the ground. Assume that the acceleration due to gravity is 9.81 m divided by sec squared and let​ x(t) represent the distance the object has fallen in t seconds. find the time it takes to hit the ground

We can model the motion of the object using Newton's second law of motion, which is given by:

\[
F = ma
\]
where \( F \) is the total force acting on the object, \( m \) is the mass of the object, and \( a \) is the acceleration of the object.

### Step 1: Set up the forces

- The gravitational force on the object is:
  \[
  F_{\text{gravity}} = mg = 5 \times 9.81 = 49.05 \, \text{N}
  \]

- The force due to air resistance is given by \( F_{\text{air}} = -20v \), where \( v(t) \) is the velocity of the object at time \( t \).

Thus, the total force acting on the object is:
\[
F = F_{\text{gravity}} + F_{\text{air}} = 49.05 - 20v
\]

### Step 2: Apply Newton's second law

Using Newton's second law \( F = ma \), and noting that \( a = \frac{dv}{dt} \), we get:
\[
49.05 - 20v = 5 \frac{dv}{dt}
\]
Simplifying, we get the differential equation governing the velocity:
\[
\frac{dv}{dt} = 9.81 - 4v
\]

### Step 3: Solve the differential equation for velocity

This is a first-order linear differential equation. We solve it using an integrating factor:

The equation is of the form:
\[
\frac{dv}{dt} + 4v = 9.81
\]

The integrating factor is \( e^{4t} \). Multiply both sides of the equation by \( e^{4t} \):
\[
e^{4t} \frac{dv}{dt} + 4e^{4t}v = 9.81 e^{4t}
\]

Now the left-hand side is the derivative of \( e^{4t}v \), so we can integrate both sides:
\[
\frac{d}{dt} (e^{4t} v) = 9.81 e^{4t}
\]

Integrating both sides with respect to \( t \):
\[
e^{4t} v = \frac{9.81}{4} e^{4t} + C
\]

Dividing through by \( e^{4t} \):
\[
v(t) = \frac{9.81}{4} + Ce^{-4t}
\]

### Step 4: Apply the initial condition for velocity

The initial velocity is given as \( v(0) = -50 \, \text{m/s} \). Applying this condition:
\[
-50 = \frac{9.81}{4} + C
\]
\[
C = -50 - \frac{9.81}{4}
\]
\[
C = -50 - 2.4525 = -52.4525
\]

Thus, the velocity as a function of time is:
\[
v(t) = \frac{9.81}{4} - 52.4525 e^{-4t}
\]

### Step 5: Find the equation of motion

To find the distance fallen \( x(t) \), we integrate the velocity function:
\[
x(t) = \int v(t) \, dt = \int \left( \frac{9.81}{4} - 52.4525 e^{-4t} \right) dt
\]

The integral is:
\[
x(t) = \frac{9.81}{4} t + \frac{52.4525}{4} e^{-4t} + D
\]

### Step 6: Apply the initial condition for position

The object starts 400 m above the ground, so \( x(0) = 0 \). Applying this:
\[
0 = 0 + \frac{52.4525}{4} + D
\]
\[
D = - \frac{52.4525}{4} = -13.1131
\]

Thus, the position as a function of time is:
\[
x(t) = \frac{9.81}{4} t + 13.1131 \left( 1 - e^{-4t} \right)
\]

### Step 7: Determine when the object strikes the ground

The object strikes the ground when it has fallen 400 meters, so set \( x(t) = 400 \) and solve for \( t \):
\[
400 = \frac{9.81}{4} t + 13.1131 \left( 1 - e^{-4t} \right)
\]

Let me solve this equation for \( t \).The object will strike the ground approximately after 157.75 seconds.

Would you like more details on any of the steps or have further questions? 

Here are 5 related questions you might find interesting:
1. How does air resistance affect the time it takes for the object to hit the ground compared to no air resistance?
2. What would happen if the object was dropped from a higher altitude, like 1000 meters?
3. How would the equation of motion change if the air resistance was proportional to \(v^2\) instead of \(v\)?
4. How can you derive the terminal velocity from the velocity equation?
5. How would the motion be affected if gravity varied with altitude?

**Tip:** To solve differential equations involving forces like gravity and air resistance, always set up the forces clearly and apply Newton's second law for easier integration.

An object of mass 5 kg is given an initial downward velocity of 50 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is minus20v, where v is the velocity of the object in m/sec. Determine the equation of motion of the object. If the object is initially 400 m above the ground, determine when the object will strike the ground. Assume that the acceleration due to gravity is 9.81 m/sec^2 and let x(t) represent the distance the object has fallen in t seconds. Find the time it takes to hit the ground.

Learn how to determine the equation of motion for an object of mass 5 kg falling under gravity with air resistance. This detailed step-by-step solution uses Newton's second law and first-order linear differential equations. Explore velocity and position functions, and calculate when the object will hit the ground.

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