A particle passes a given point moving downwards with a velocity of fifty metres per second. How long before this was moving upwards at the same rate.

Answer

To solve the problem, we need to determine how long before the particle was moving upwards at a velocity of 50 m/s, given that it is currently moving downwards at the same velocity.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle is currently moving downwards with a velocity of \( v = -50 \, \text{m/s} \) (we take downward as negative).
- We need to find the time \( t \) before this point when the particle was moving upwards at the same speed, which means it was moving with a velocity of \( u = +50 \, \text{m/s} \) (upward is positive).

2. **Acceleration**:
- The particle is under the influence of gravity, which causes it to accelerate downwards. The acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \) (downward).

3. **Using the Equation of Motion**:
- We can use the first equation of motion:
\[ v = u + at \]
- Here, \( v \) is the final velocity (downward), \( u \) is the initial velocity (upward), \( a \) is the acceleration (downward), and \( t \) is the time interval we want to find.

4. **Substituting the Values**:
- Substitute \( v = -50 \, \text{m/s} \), \( u = 50 \, \text{m/s} \), and \( a = -10 \, \text{m/s}^2 \) into the equation:
\[ -50 = 50 + (-10)t \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ -50 - 50 = -10t \]
\[ -100 = -10t \]

6. **Solving for Time \( t \)**:
- Dividing both sides by -10:
\[ t = \frac{100}{10} = 10 \, \text{seconds} \]

### Final Answer:
The particle was moving upwards at a velocity of 50 m/s, 10 seconds before it started moving downwards at the same velocity.

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