A stone is dropped from a height `h` . It hits the ground with a certain momentum `P` . If the same stone is dropped from a height `100%` more thanthe preyiious height, the momentum when it hits the ground will change by

Options

A. `41%`
B. `200%`
C. `100%`
D. `68%`

Explanation

To solve the problem step by step, we will analyze the situation where a stone is dropped from a height and calculate the change in momentum when the height is increased by 100%.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A stone is dropped from a height \( h \) and hits the ground with a momentum \( P_1 \).
- The stone is then dropped from a height that is 100% more than the previous height, which means the new height \( h_2 = h + h = 2h \).

2. **Momentum Calculation**:
- The momentum \( P \) of an object is given by the formula:
\[ P = m \cdot v \]
where \( m \) is the mass of the stone and \( v \) is its velocity just before hitting the ground.

3. **Finding the Velocity for Height \( h \)**:
- Using the kinematic equation for an object in free fall:
\[ v^2 = u^2 + 2gh \]
where \( u = 0 \) (initial velocity when dropped), we have:
\[ v^2 = 0 + 2gh \implies v = \sqrt{2gh} \]
- Therefore, the initial momentum \( P_1 \) when dropped from height \( h \) is:
\[ P_1 = m \cdot \sqrt{2gh} \]

4. **Finding the Velocity for Height \( 2h \)**:
- Now, for the new height \( h_2 = 2h \):
\[ v^2 = u^2 + 2gh_2 = 0 + 2g(2h) = 4gh \implies v = \sqrt{4gh} = 2\sqrt{gh} \]
- The new momentum \( P_2 \) when dropped from height \( 2h \) is:
\[ P_2 = m \cdot 2\sqrt{gh} \]

5. **Relating \( P_2 \) to \( P_1 \)**:
- We can express \( P_2 \) in terms of \( P_1 \):
\[ P_2 = 2 \cdot \sqrt{2} \cdot P_1 = \sqrt{2} \cdot P_1 \]

6. **Calculating the Change in Momentum**:
- The change in momentum is given by:
\[ \Delta P = P_2 - P_1 = \sqrt{2} \cdot P_1 - P_1 = (\sqrt{2} - 1) \cdot P_1 \]

7. **Calculating Percentage Change in Momentum**:
- The percentage change in momentum is calculated as:
\[ \text{Percentage Change} = \frac{\Delta P}{P_1} \cdot 100 = \frac{(\sqrt{2} - 1) \cdot P_1}{P_1} \cdot 100 = (\sqrt{2} - 1) \cdot 100 \]
- Substituting \( \sqrt{2} \approx 1.414 \):
\[ \text{Percentage Change} = (1.414 - 1) \cdot 100 = 0.414 \cdot 100 = 41.4\% \]

8. **Final Answer**:
- The percentage change in momentum when the stone is dropped from a height that is 100% more than the previous height is approximately **41.4%**.