To solve the problem step-by-step, we will follow the concepts of thrust, mass flow rate, and the equations of motion.
### Step 1: Identify the given data
- Initial mass of the rocket, \( m_0 = 1000 \, \text{kg} \)
- Rate of fuel consumption, \( \frac{dm}{dt} = -10 \, \text{kg/s} \) (negative because mass is decreasing)
- Relative velocity of the ejected fuel, \( V_r = 2000 \, \text{m/s} \)
- Time of burning fuel, \( t = 60 \, \text{s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
### Step 2: Calculate the final mass of the rocket after burning fuel
The mass of the rocket decreases as it burns fuel. The final mass \( m \) after burning for 60 seconds can be calculated as:
\[
m = m_0 - k \cdot t
\]
Substituting the values:
\[
m = 1000 \, \text{kg} - (10 \, \text{kg/s} \cdot 60 \, \text{s}) = 1000 \, \text{kg} - 600 \, \text{kg} = 400 \, \text{kg}
\]
### Step 3: Write the equation for the maximum velocity of the rocket
The maximum velocity \( V \) of the rocket can be derived from the thrust and the effects of gravity. The equation is:
\[
V = V_0 - g t + V_r \ln\left(\frac{m_0}{m}\right)
\]
Where:
- \( V_0 \) is the initial velocity (0 m/s since it starts from rest)
- \( g \) is the acceleration due to gravity
- \( t \) is the time of fuel burning
- \( V_r \) is the relative velocity of the ejected fuel
- \( m_0 \) is the initial mass
- \( m \) is the final mass after burning fuel
### Step 4: Substitute the values into the equation
Now substituting the known values into the equation:
\[
V = 0 - (10 \, \text{m/s}^2 \cdot 60 \, \text{s}) + 2000 \, \text{m/s} \cdot \ln\left(\frac{1000 \, \text{kg}}{400 \, \text{kg}}\right)
\]
Calculating each term:
1. The gravitational term:
\[
-g t = -10 \cdot 60 = -600 \, \text{m/s}
\]
2. The logarithmic term:
\[
\frac{m_0}{m} = \frac{1000}{400} = 2.5
\]
\[
\ln(2.5) \approx 0.9163
\]
3. The thrust contribution:
\[
V_r \ln\left(\frac{m_0}{m}\right) = 2000 \cdot 0.9163 \approx 1832.6 \, \text{m/s}
\]
### Step 5: Calculate the maximum velocity
Combining all the terms:
\[
V = -600 + 1832.6 \approx 1232.6 \, \text{m/s}
\]
### Final Answer
The maximum velocity of the rocket is approximately \( V \approx 1232.6 \, \text{m/s} \).
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