A rocket of initial mass `m_0` has a mass `m_0(1-t//3)` at time t. The rocket is lauched from rest vertically upwards under gravity and expels burnt fuel at a speed u relative to the rocket vertically downward. Find the speed of rocket at `t=1`.

Explanation

To solve the problem, we need to find the speed of the rocket at time \( t = 1 \) second. We will use the principles of conservation of momentum and the given mass function of the rocket.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The initial mass of the rocket is \( m_0 \).
- The mass of the rocket at time \( t \) is given by \( m(t) = m_0 \left(1 - \frac{t}{3}\right) \).
- The rocket is launched from rest, so the initial velocity \( v_0 = 0 \).

2. **Determine the Mass at \( t = 1 \)**:
- Substitute \( t = 1 \) into the mass function:
\[ m(1) = m_0 \left(1 - \frac{1}{3}\right) = m_0 \left(\frac{2}{3}\right) = \frac{2m_0}{3} \]

3. **Relative Speed of Fuel Expulsion**:
- The speed of the expelled fuel relative to the rocket is \( u \) downward. Therefore, the speed of the expelled fuel with respect to the ground is \( v_r = v - u \), where \( v \) is the speed of the rocket.

4. **Apply Conservation of Momentum**:
- The momentum before and after the fuel is expelled must be equal. Initially, the momentum is \( m_0 \cdot 0 = 0 \).
- After expelling a small amount of fuel \( dm \), the momentum is:
\[ m(t) v + dm (v - u) = 0 \]
- Rearranging gives:
\[ m(t) v = -dm (v - u) \]

5. **Express \( dm \)**:
- The change in mass \( dm \) over a small time \( dt \) can be expressed as:
\[ dm = -\frac{m_0}{3} dt \]
- This is derived from the mass function since the mass decreases linearly with time.

6. **Substitute \( dm \) into the Momentum Equation**:
- Substitute \( m(t) \) and \( dm \):
\[ \frac{2m_0}{3} v = -\left(-\frac{m_0}{3}\right) (v - u) \]
- Simplifying gives:
\[ \frac{2m_0}{3} v = \frac{m_0}{3} (v - u) \]

7. **Solve for \( v \)**:
- Cancel \( m_0/3 \) from both sides:
\[ 2v = v - u \]
- Rearranging gives:
\[ v + u = 0 \implies v = -u \]

8. **Calculate the Speed at \( t = 1 \)**:
- Substitute \( t = 1 \) into the expression for speed:
\[ v = -u \ln\left(\frac{3}{2}\right) - g \]
- The final expression for the speed of the rocket at \( t = 1 \) second is:
\[ v = u \ln\left(\frac{3}{2}\right) - g \]

### Final Result:
The speed of the rocket at \( t = 1 \) second is:
\[ v = u \ln\left(\frac{3}{2}\right) - g \]