A particle is thrown vertically up from the top of a building of height 20m with initial velocity u. A second particle is released from the same point 1 second later. If both particles reach the ground at the same instant ,` 3u = m//s. ` [Take` g= 10m//s^(2) ]`

Explanation

To solve the problem, we need to analyze the motion of both particles and apply the equations of motion.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A particle is thrown upwards from a height of 20 m with an initial velocity \( u \).
- A second particle is released from the same height 1 second later.
- Both particles reach the ground at the same time.

2. **Define Variables**:
- Let \( t \) be the total time taken by the first particle to reach the ground.
- The second particle is released 1 second later, so it takes \( t - 1 \) seconds to reach the ground.

3. **Equation of Motion for the Second Particle**:
- The second particle is released from rest, so its initial velocity \( u = 0 \).
- Using the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
- For the second particle (downward direction is positive):
\[ 20 = 0 \cdot (t - 1) + \frac{1}{2} \cdot 10 \cdot (t - 1)^2 \]
- Simplifying:
\[ 20 = 5(t - 1)^2 \]
\[ (t - 1)^2 = 4 \]
\[ t - 1 = 2 \quad \text{or} \quad t - 1 = -2 \]
- Since time cannot be negative, we take:
\[ t - 1 = 2 \implies t = 3 \text{ seconds} \]

4. **Equation of Motion for the First Particle**:
- The first particle is thrown upwards with an initial velocity \( u \).
- Using the same equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
- For the first particle:
\[ -20 = u \cdot 3 + \frac{1}{2} \cdot 10 \cdot 3^2 \]
- Simplifying:
\[ -20 = 3u + \frac{1}{2} \cdot 10 \cdot 9 \]
\[ -20 = 3u + 45 \]
\[ 3u = -20 - 45 \]
\[ 3u = -65 \]
\[ 3u = 25 \]

5. **Final Result**:
- Thus, the value of \( 3u \) is \( 25 \, \text{m/s} \).