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. `250sqrt(2)` m

Explanation

To solve the problem step by step, we will analyze the motion of the two bodies thrown from the towers.

### Step 1: Analyze the first body
The first body of mass \( m \) is thrown horizontally with velocity \( v \) from a height \( h \). The horizontal distance it covers when it touches the ground is given as \( 250 \, m \).

The time of flight \( t_1 \) for the first body can be calculated using the formula for free fall:
\[ t_1 = \sqrt{\frac{2h}{g}} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Calculate the horizontal distance for the first body
The horizontal distance \( x_1 \) covered by the first body can be expressed as:
\[ x_1 = v \cdot t_1 \]
Substituting \( t_1 \):
\[ x_1 = v \cdot \sqrt{\frac{2h}{g}} \]
Given that \( x_1 = 250 \, m \), we can write:
\[ 250 = v \cdot \sqrt{\frac{2h}{g}} \quad \text{(1)} \]

### Step 3: Analyze the second body
The second body of mass \( 2m \) is thrown horizontally with velocity \( \frac{v}{2} \) from a height of \( 4h \). We need to find the horizontal distance \( x \) it covers when it touches the ground.

### Step 4: Calculate the time of flight for the second body
The time of flight \( t_2 \) for the second body can be calculated similarly:
\[ t_2 = \sqrt{\frac{2(4h)}{g}} = \sqrt{\frac{8h}{g}} \]

### Step 5: Calculate the horizontal distance for the second body
The horizontal distance \( x \) covered by the second body can be expressed as:
\[ x = \left(\frac{v}{2}\right) \cdot t_2 \]
Substituting \( t_2 \):
\[ x = \left(\frac{v}{2}\right) \cdot \sqrt{\frac{8h}{g}} \]

### Step 6: Relate \( x \) to \( x_1 \)
Now we can express \( x \) in terms of \( x_1 \) using equation (1):
\[ x = \left(\frac{v}{2}\right) \cdot \sqrt{\frac{8h}{g}} = \frac{v}{2} \cdot 2 \cdot \sqrt{\frac{2h}{g}} = v \cdot \sqrt{\frac{2h}{g}} \]
This implies:
\[ x = 2 \cdot \left(\frac{v}{2} \cdot \sqrt{\frac{2h}{g}}\right) \]
From equation (1), we know that:
\[ v \cdot \sqrt{\frac{2h}{g}} = 250 \]
Thus:
\[ x = 2 \cdot 125 = 250 \]

### Final Result
The value of \( x \) is \( 250 \, m \).