From top of a tower of hight 80m, person drops a stone. And after one second another stone is thrown down from the top of the same tower with some initial velocity, such that both the stones reach the ground at the same instant. Calculate the magnitude of the initial velociy (in m/s) of the second stone. (Take `g=10m//s^(2)`)

Explanation

To solve the problem, we will analyze the motion of both stones dropped from the tower step by step.

### Given:
- Height of the tower, \( h = 80 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- The first stone is dropped (initial velocity \( u_1 = 0 \)).
- The second stone is thrown down after 1 second with an initial velocity \( u_2 \).

### Step 1: Calculate the time taken for the first stone to reach the ground.
The first stone is dropped from rest, so we can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
For the first stone:
- \( s = 80 \, \text{m} \)
- \( u = 0 \)
- \( a = g = 10 \, \text{m/s}^2 \)

Substituting the values:
\[ 80 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ 80 = 5t^2 \]
\[ t^2 = \frac{80}{5} = 16 \]
\[ t = \sqrt{16} = 4 \, \text{s} \]
Thus, the first stone takes 4 seconds to reach the ground.

### Step 2: Determine the time taken for the second stone to reach the ground.
Since the second stone is thrown 1 second after the first stone, it will take:
\[ t_2 = t_1 - 1 = 4 - 1 = 3 \, \text{s} \]

### Step 3: Use the equation of motion for the second stone.
For the second stone, which is thrown down with an initial velocity \( u_2 \):
\[ s = u_2 t + \frac{1}{2} a t^2 \]
Substituting the values:
- \( s = 80 \, \text{m} \)
- \( t = 3 \, \text{s} \)
- \( a = g = 10 \, \text{m/s}^2 \)

The equation becomes:
\[ 80 = u_2 \cdot 3 + \frac{1}{2} \cdot 10 \cdot (3^2) \]
\[ 80 = 3u_2 + \frac{1}{2} \cdot 10 \cdot 9 \]
\[ 80 = 3u_2 + 45 \]

### Step 4: Solve for \( u_2 \).
Rearranging the equation:
\[ 3u_2 = 80 - 45 \]
\[ 3u_2 = 35 \]
\[ u_2 = \frac{35}{3} \approx 11.67 \, \text{m/s} \]

### Final Answer:
The magnitude of the initial velocity of the second stone is approximately \( 11.67 \, \text{m/s} \).

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