A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is m, the fuel is consumed at rate r, and the exhaust gases are ejected with constant velocity ve (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation
v(t) = −gt − ve ln 
m − rt
m
where g is the acceleration due to gravity and t is not too large. If g = 9.8 m/s2, m = 27,000 kg, r = 130 kg/s, and ve = 2,900 m/s, find the height of the rocket one minute after liftoff. (Round your answer to the nearest whole meter.)

We are given the velocity equation for the rocket as:

\[
v(t) = -gt - v_e \ln\left(\frac{m - rt}{m}\right)
\]

where:
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( m = 27,000 \, \text{kg} \) (initial mass of the rocket),
- \( r = 130 \, \text{kg/s} \) (rate of fuel consumption),
- \( v_e = 2,900 \, \text{m/s} \) (velocity of the ejected gases),
- \( t = 60 \, \text{seconds} \) (since we are calculating the height after one minute).

We need to find the height of the rocket at \( t = 60 \) seconds. The height of the rocket is the integral of the velocity over time:

\[
h(t) = \int v(t) \, dt
\]

### Step 1: Substitute values into the velocity equation
First, substitute the known values into the velocity equation:

\[
v(t) = -9.8t - 2,900 \ln\left(\frac{27,000 - 130t}{27,000}\right)
\]

### Step 2: Integrate the velocity equation
Now, to find the height, we need to integrate this velocity function from \( t = 0 \) to \( t = 60 \).

\[
h(60) = \int_0^{60} \left[ -9.8t - 2,900 \ln\left(\frac{27,000 - 130t}{27,000}\right) \right] dt
\]

I will compute this integral now.The height expression after integrating the velocity function is:

\[
h(60) = -48982.71 - 174000 \cdot \ln\left(\frac{32}{45}\right)
\]

I will now evaluate this expression numerically to find the rocket's height at \( t = 60 \) seconds.The height of the rocket after 60 seconds (one minute) is approximately 10,339 meters.

Would you like further details or clarifications? 

Here are 5 related questions:
1. How does fuel consumption rate affect the rocket's velocity and height?
2. What happens to the rocket's height if the exhaust velocity increases?
3. How would a different gravitational acceleration impact the results?
4. Can we model the rocket's height after its fuel is exhausted?
5. How can we compute the time at which the rocket reaches its maximum height?

**Tip:** Integrating velocity functions is a standard method for calculating displacement or height when velocity changes over time.

A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is m, the fuel is consumed at rate r, and the exhaust gases are ejected with constant velocity ve (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation v(t) = −gt − ve ln (m − rt/m) where g is the acceleration due to gravity and t is not too large. If g = 9.8 m/s2, m = 27,000 kg, r = 130 kg/s, and ve = 2,900 m/s, find the height of the rocket one minute after liftoff. (Round your answer to the nearest whole meter.)

Learn how to calculate the height of a rocket one minute after liftoff, given fuel consumption, exhaust velocity, and gravitational acceleration. This problem involves integrating the velocity function derived from the changing mass of the rocket. Suitable for university-level physics or engineering students.

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