A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone

Options

A. `12.25 m//s`
B. `14.75 m//s`
C. `16.23 m//s`
D. `17.15 m//s`

Answer

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

### Step 1: Determine the time taken by the first stone
The first stone is dropped from a height of 44.1 m. Since it is dropped, its initial velocity (u) is 0 m/s. We can use the second equation of motion:

\[ h = ut + \frac{1}{2}gt^2 \]

Where:
- \( h = 44.1 \, \text{m} \)
- \( u = 0 \, \text{m/s} \)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t \) is the time taken to reach the water.

Substituting the values into the equation:

\[ 44.1 = 0 + \frac{1}{2} \times 9.8 \times t^2 \]

This simplifies to:

\[ 44.1 = 4.9t^2 \]

### Step 2: Solve for \( t^2 \)
Rearranging gives:

\[ t^2 = \frac{44.1}{4.9} \]

Calculating this:

\[ t^2 = 9 \quad \Rightarrow \quad t = 3 \, \text{s} \]

So, the first stone takes 3 seconds to hit the water.

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

\[ t' = t - 1 = 3 - 1 = 2 \, \text{s} \]

### Step 4: Use the second stone's motion equation
For the second stone, which is thrown downward with an initial velocity \( u \), we use the same equation of motion:

\[ h = ut' + \frac{1}{2}gt'^2 \]

Substituting the known values:

\[ 44.1 = u \cdot 2 + \frac{1}{2} \cdot 9.8 \cdot (2^2) \]

This simplifies to:

\[ 44.1 = 2u + \frac{1}{2} \cdot 9.8 \cdot 4 \]

Calculating the second term:

\[ \frac{1}{2} \cdot 9.8 \cdot 4 = 19.6 \]

So we have:

\[ 44.1 = 2u + 19.6 \]

### Step 5: Solve for \( u \)
Rearranging gives:

\[ 2u = 44.1 - 19.6 \]

Calculating the right side:

\[ 2u = 24.5 \quad \Rightarrow \quad u = \frac{24.5}{2} = 12.25 \, \text{m/s} \]

### Final Answer
The initial speed of the second stone is \( 12.25 \, \text{m/s} \).

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