A boy on a cliff 49 m high drops a stone. One second later, he throws a second stone after the first. They both hit the ground at the same time. With what speed did he second stone?

Explanation

To solve the problem, we need to analyze the motion of both stones dropped by the boy from the cliff. Here's a step-by-step solution:

### Step 1: Calculate the time taken for the first stone to hit the ground.

The first stone is dropped from a height of 49 m. We can use the equation of motion for free fall:

\[ H = UT + \frac{1}{2} g T^2 \]

Where:
- \( H = 49 \, \text{m} \) (height)
- \( U = 0 \, \text{m/s} \) (initial velocity, since it is dropped)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( T \) is the time taken to hit the ground.

Substituting the values:

\[ 49 = 0 + \frac{1}{2} \times 9.8 \times T^2 \]

This simplifies to:

\[ 49 = 4.9 T^2 \]

Now, solving for \( T^2 \):

\[ T^2 = \frac{49}{4.9} = 10 \]

Taking the square root:

\[ T = \sqrt{10} \approx 3.162 \, \text{s} \]

### Step 2: Determine the time taken for the second stone.

The second stone is thrown 1 second after the first stone is dropped. Therefore, the time taken for the second stone to hit the ground is:

\[ T_{second} = T - 1 = 3.162 - 1 = 2.162 \, \text{s} \]

### Step 3: Apply the equation of motion for the second stone.

For the second stone, we use the same equation of motion:

\[ H = UT + \frac{1}{2} g T^2 \]

Where:
- \( H = 49 \, \text{m} \)
- \( U \) is the initial velocity of the second stone (which we need to find).
- \( g = 9.8 \, \text{m/s}^2 \)
- \( T = 2.162 \, \text{s} \)

Substituting the values:

\[ 49 = U \times 2.162 + \frac{1}{2} \times 9.8 \times (2.162)^2 \]

Calculating \( \frac{1}{2} \times 9.8 \times (2.162)^2 \):

\[ \frac{1}{2} \times 9.8 \times 4.676244 = 22.90 \, \text{m} \]

So the equation becomes:

\[ 49 = U \times 2.162 + 22.90 \]

### Step 4: Solve for \( U \).

Rearranging the equation:

\[ U \times 2.162 = 49 - 22.90 \]

Calculating the right side:

\[ U \times 2.162 = 26.10 \]

Now, solving for \( U \):

\[ U = \frac{26.10}{2.162} \approx 12.1 \, \text{m/s} \]

### Final Answer:

The speed with which the second stone was thrown is approximately **12.1 m/s**.

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