A stone is dropped into a well of 20 m deep. Another stone is thrown downward with velocity 'v' one second later. If both stones reach the water surface in the well simultan eously, v is equal to `(g = 10 ms^(-2))`

Options

A. 30 `"ms"^(-1)`
B. 15 `"ms"^(-1)`
C. 20 `"ms"^(-1)`
D. 10 `"ms"^(-1)`

Answer

To solve the problem step by step, we need to analyze the motion of both stones dropped into the well.

### Step 1: Understand the problem
We have two stones:
- The first stone is dropped from the top of a well that is 20 meters deep.
- The second stone is thrown downward with an initial velocity \( v \) one second after the first stone is dropped.
- Both stones reach the water surface simultaneously.

### Step 2: Determine the time taken by the first stone
The first stone is dropped, so its initial velocity \( u_1 = 0 \) m/s. We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s = 20 \) m (depth of the well)
- \( u = 0 \) m/s (initial velocity of the first stone)
- \( a = g = 10 \) m/s² (acceleration due to gravity)

Substituting these values into the equation:
\[ 20 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 20 = 5t^2 \]
Dividing both sides by 5:
\[ t^2 = 4 \]
Taking the square root:
\[ t = 2 \text{ seconds} \]
So, the first stone takes 2 seconds to reach the water.

### Step 3: Determine the time taken by the second stone
The second stone is thrown downward 1 second after the first stone. Therefore, it has only 1 second to reach the water surface.

### Step 4: Use the equation of motion for the second stone
For the second stone, we can use the same equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s = 20 \) m (depth of the well)
- \( u = v \) (initial velocity of the second stone, which we need to find)
- \( t = 1 \) s (time taken by the second stone to reach the water)
- \( a = g = 10 \) m/s² (acceleration due to gravity)

Substituting these values into the equation:
\[ 20 = v \cdot 1 + \frac{1}{2} \cdot 10 \cdot (1)^2 \]
This simplifies to:
\[ 20 = v + 5 \]
Rearranging gives:
\[ v = 20 - 5 \]
Thus:
\[ v = 15 \text{ m/s} \]

### Final Answer
The initial velocity \( v \) of the second stone is \( 15 \) m/s.

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