A body of mass m thrown horizontally with velocity v, from the top of tower of height h touches the level ground at a distance of 250m from the foot of the tower. A body of mass 2m thrown horizontally with velocity `v//2`, from the top of tower of height 4h will touch the level ground at a distance x from the foot of tower. The value of x is

Options

A. 250 m
B. 500 m
C. 125 m
D. `250sqrt2` m

Answer

To solve the problem, we will analyze the motion of both bodies thrown from the tower and use the formula for the range of horizontal projectile motion.

### Step-by-Step Solution:

1. **Understanding the First Body's Motion**:
- A body of mass \( m \) is thrown horizontally with an initial velocity \( v \) from a height \( h \).
- The range \( R_1 \) (horizontal distance traveled) is given as 250 m.
- The formula for the range of a horizontally thrown projectile is:
\[ R = u \cdot \sqrt{\frac{2h}{g}} \]
- Here, \( u = v \), \( h = h \), and \( g \) is the acceleration due to gravity.
- Therefore, we can write:
\[ 250 = v \cdot \sqrt{\frac{2h}{g}} \quad \text{(1)} \]

2. **Understanding the Second Body's Motion**:
- A second body of mass \( 2m \) is thrown horizontally with an initial velocity of \( \frac{v}{2} \) from a height of \( 4h \).
- Let the range for this body be \( x \).
- Using the same formula for the range:
\[ x = \left(\frac{v}{2}\right) \cdot \sqrt{\frac{2(4h)}{g}} \]
- Simplifying this expression:
\[ x = \left(\frac{v}{2}\right) \cdot \sqrt{\frac{8h}{g}} \]
\[ x = \left(\frac{v}{2}\right) \cdot 2 \cdot \sqrt{\frac{2h}{g}} \]
\[ x = v \cdot \sqrt{\frac{2h}{g}} \quad \text{(2)} \]

3. **Relating Equations (1) and (2)**:
- From equation (1), we have:
\[ v \cdot \sqrt{\frac{2h}{g}} = 250 \]
- Substituting this into equation (2):
\[ x = 250 \]

4. **Conclusion**:
- Therefore, the value of \( x \) is:
\[ x = 250 \text{ m} \]

### Final Answer:
The value of \( x \) is **250 m**.