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 step by step, we will analyze the motion of both particles and apply the equations of motion.

### Step 1: Understand the problem
We have two particles:
- Particle 1 is thrown vertically upwards from the top of a 20 m building with an initial velocity \( u \).
- Particle 2 is released from the same point 1 second later.

Both particles reach the ground at the same time.

### Step 2: Define the time of flight for both particles
Let the time taken by Particle 1 to reach the ground be \( T \). Therefore, the time taken by Particle 2 to reach the ground will be \( T - 1 \) seconds (since it is released 1 second later).

### Step 3: Write the equation of motion for Particle 1
Using the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
For Particle 1:
- Displacement \( S = -20 \) m (downward)
- Initial velocity \( u \)
- Acceleration \( a = -g = -10 \, \text{m/s}^2 \) (acting downward)

The equation becomes:
\[ -20 = uT - \frac{1}{2} \cdot 10 \cdot T^2 \]
This simplifies to:
\[ -20 = uT - 5T^2 \quad \text{(1)} \]

### Step 4: Write the equation of motion for Particle 2
For Particle 2:
- Displacement \( S = -20 \) m (downward)
- Initial velocity = 0 (since it is released)
- Acceleration \( a = -g = -10 \, \text{m/s}^2 \)

The equation becomes:
\[ -20 = 0 \cdot (T - 1) - \frac{1}{2} \cdot 10 \cdot (T - 1)^2 \]
This simplifies to:
\[ -20 = -5(T - 1)^2 \]
Dividing by -5 gives:
\[ 4 = (T - 1)^2 \]
Taking the square root:
\[ T - 1 = 2 \quad \text{or} \quad T - 1 = -2 \]
Thus, \( T = 3 \) seconds (we ignore the negative solution as time cannot be negative).

### Step 5: Substitute \( T \) back into the equation for Particle 1
Now substituting \( T = 3 \) seconds into equation (1):
\[ -20 = u(3) - 5(3^2) \]
This simplifies to:
\[ -20 = 3u - 45 \]
Rearranging gives:
\[ 3u = 45 - 20 \]
\[ 3u = 25 \]

### Step 6: Solve for \( u \)
Now we can find \( u \):
\[ u = \frac{25}{3} \, \text{m/s} \]

### Final Answer
Thus, the initial velocity \( u \) is:
\[ u = \frac{25}{3} \, \text{m/s} \]