A time `t = 0`, a body of mass `1 kg` is dropped from height `500 m`. At `t = 1 sec`, the body explodes into two parts of equal mass. One of them moves vertically upwards with velocity `5 m//s` just after the explosion. The distance (in `m`) between the parts at time `t = 4 sec` is `15x` Then find `x`?

Explanation

To solve the problem, we will break it down into steps.

### Step 1: Determine the velocity of the body just before the explosion
The body is dropped from a height of 500 m. The initial velocity (u) is 0 m/s, and the acceleration due to gravity (a) is approximately 10 m/s². We need to find the velocity (v) after 1 second.

Using the equation of motion:
\[ v = u + at \]
\[ v = 0 + (10 \, \text{m/s}^2)(1 \, \text{s}) \]
\[ v = 10 \, \text{m/s} \]

### Step 2: Analyze the explosion
At \( t = 1 \) second, the body explodes into two parts of equal mass (0.5 kg each). One part moves vertically upwards with a velocity of 5 m/s just after the explosion.

Let’s denote:
- Mass of the first part moving upwards = 0.5 kg
- Velocity of the first part (upwards) = 5 m/s
- Mass of the second part moving downwards = 0.5 kg
- Velocity of the second part (downwards) = \( V \) m/s (to be determined)

### Step 3: Apply conservation of momentum
Using the conservation of momentum before and after the explosion:
Before the explosion:
- Total momentum = mass × velocity = \( 1 \, \text{kg} \times 10 \, \text{m/s} = 10 \, \text{kg m/s} \)

After the explosion:
- Total momentum = momentum of first part + momentum of second part
\[ 10 = (0.5 \times 5) + (0.5 \times V) \]
\[ 10 = 2.5 + 0.5V \]
\[ 0.5V = 10 - 2.5 \]
\[ 0.5V = 7.5 \]
\[ V = \frac{7.5}{0.5} = 15 \, \text{m/s} \]

### Step 4: Determine the distance traveled by each part after the explosion
**For the part moving upwards:**
- Initial velocity = 5 m/s
- Time = 3 seconds (from \( t = 1 \) to \( t = 4 \))
- Acceleration = -10 m/s² (due to gravity)

Using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
\[ s_{up} = 5(3) + \frac{1}{2}(-10)(3^2) \]
\[ s_{up} = 15 - 45 = -30 \, \text{m} \]
(The negative sign indicates it has moved 30 m upwards from the point of explosion.)

**For the part moving downwards:**
- Initial velocity = 15 m/s
- Time = 3 seconds
- Acceleration = 10 m/s²

Using the same equation:
\[ s_{down} = ut + \frac{1}{2} a t^2 \]
\[ s_{down} = 15(3) + \frac{1}{2}(10)(3^2) \]
\[ s_{down} = 45 + 45 = 90 \, \text{m} \]

### Step 5: Calculate the total distance between the two parts
The distance between the two parts at \( t = 4 \) seconds is:
\[ \text{Distance} = s_{down} - (-s_{up}) \]
\[ \text{Distance} = 90 - (-30) = 90 + 30 = 120 \, \text{m} \]

### Step 6: Relate the distance to 15x
According to the problem, this distance is given as \( 15x \):
\[ 120 = 15x \]
\[ x = \frac{120}{15} = 8 \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{8} \]