A rock is dropped from rest from a height h above the ground. It falls and hits the ground with a speed of 11 m/s. From what height should the rock be dropped so that its speed on hitting the ground is 22 m/s ? Neglect air resistance.

Options

A. 1.4 h
B. 3.0 h
C. 2.0 h
D. 4.0 h

Answer

To solve the problem, we need to determine the height from which a rock should be dropped so that it hits the ground with a speed of 22 m/s, given that it hits the ground with a speed of 11 m/s when dropped from a certain height \( h \).

### Step-by-Step Solution:

1. **Understanding the Motion**:
When an object is dropped from rest, it accelerates downward due to gravity. The final velocity \( v \) of the object just before it hits the ground can be calculated using the equation of motion:
\[ v^2 = u^2 + 2as \]
where \( u \) is the initial velocity (0 m/s for a dropped object), \( a \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( s \) is the distance fallen (height).

2. **Applying the Equation for the First Case**:
For the first case where the rock hits the ground with a speed of 11 m/s, we can set up the equation:
\[ (11)^2 = (0)^2 + 2 \cdot g \cdot h_1 \]
Simplifying this gives:
\[ 121 = 2gh_1 \]
Thus, we can express \( h_1 \) as:
\[ h_1 = \frac{121}{2g} \]

3. **Applying the Equation for the Second Case**:
For the second case where the rock hits the ground with a speed of 22 m/s, we set up a similar equation:
\[ (22)^2 = (0)^2 + 2 \cdot g \cdot h_2 \]
Simplifying gives:
\[ 484 = 2gh_2 \]
Thus, we can express \( h_2 \) as:
\[ h_2 = \frac{484}{2g} \]

4. **Finding the Relationship Between Heights**:
Now we have two expressions for \( h_1 \) and \( h_2 \):
\[ h_1 = \frac{121}{2g} \quad \text{and} \quad h_2 = \frac{484}{2g} \]
To find the relationship between \( h_2 \) and \( h_1 \):
\[ h_2 = 4h_1 \]

5. **Conclusion**:
Therefore, the height \( h_2 \) from which the rock should be dropped to hit the ground with a speed of 22 m/s is four times the height \( h_1 \) from which it was dropped to hit the ground with a speed of 11 m/s.

### Final Answer:
If the height from which the rock was originally dropped (to hit the ground at 11 m/s) is \( h \), then the height from which it should be dropped to hit the ground at 22 m/s is:
\[ h_2 = 4h \]