A body projected horizontally with a velocity`v` from a height `h` has a range `R`.With what velocity a body is to be projected horizontally from a height `h//2` to have the same range ?

Options

A. `sqrt2v`
B. `2v`
C. `6v`
D. `8v`

Answer

To solve the problem, we need to analyze the motion of a body projected horizontally from two different heights and determine the relationship between the velocities required for the same range.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body is projected horizontally with velocity \( v \) from a height \( h \) and has a range \( R \).
- We need to find the velocity \( v_1 \) required to project a body horizontally from a height \( \frac{h}{2} \) so that it covers the same range \( R \).

2. **Time of Flight from Height \( h \)**:
- For a body projected horizontally from height \( h \), the time of flight \( t \) can be calculated using the formula for free fall:
\[ h = \frac{1}{2} g t^2 \]
- Rearranging gives:
\[ t = \sqrt{\frac{2h}{g}} \]

3. **Range Calculation for Height \( h \)**:
- The range \( R \) when projected horizontally is given by:
\[ R = v \cdot t \]
- Substituting the expression for \( t \):
\[ R = v \cdot \sqrt{\frac{2h}{g}} \]

4. **Time of Flight from Height \( \frac{h}{2} \)**:
- Now, for the height \( \frac{h}{2} \):
\[ \frac{h}{2} = \frac{1}{2} g T^2 \]
- Rearranging gives:
\[ T = \sqrt{\frac{h}{g}} \]

5. **Range Calculation for Height \( \frac{h}{2} \)**:
- The range \( R \) when projected horizontally from height \( \frac{h}{2} \) with velocity \( v_1 \) is:
\[ R = v_1 \cdot T \]
- Substituting the expression for \( T \):
\[ R = v_1 \cdot \sqrt{\frac{h}{g}} \]

6. **Setting the Ranges Equal**:
- Since both ranges are equal, we can set the two expressions for \( R \) equal to each other:
\[ v \cdot \sqrt{\frac{2h}{g}} = v_1 \cdot \sqrt{\frac{h}{g}} \]

7. **Solving for \( v_1 \)**:
- Dividing both sides by \( \sqrt{\frac{h}{g}} \):
\[ v \cdot \sqrt{2} = v_1 \]
- Thus, we find:
\[ v_1 = v \cdot \sqrt{2} \]

### Final Answer:
The velocity \( v_1 \) required to project a body horizontally from a height \( \frac{h}{2} \) to have the same range \( R \) is:
\[ v_1 = v \sqrt{2} \]