Two bodies are thrown simultaneously from a tower with same initial velocity `v_0`: one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is

Options

A. `2v_0t+(1)/(2)gt^2`
B. `2v_0t`
C. `v_0t+(1)/(2)g` `t^2`
D. `v_0t`

Explanation

To solve the problem of finding the distance between two bodies thrown simultaneously from a tower, one vertically upwards and the other vertically downwards, we can follow these steps:

### Step 1: Define the motion of the first body (upwards)
The first body is thrown upwards with an initial velocity \( v_0 \). The position of this body after time \( t \) can be described by the equation of motion:
\[ s_1 = v_0 t - \frac{1}{2} g t^2 \]
where \( g \) is the acceleration due to gravity (acting downwards).

### Step 2: Define the motion of the second body (downwards)
The second body is thrown downwards with the same initial velocity \( v_0 \). The position of this body after time \( t \) can be described by the equation:
\[ s_2 = -v_0 t - \frac{1}{2} g t^2 \]
Here, we take the downward direction as negative, hence the negative sign in front of \( v_0 t \).

### Step 3: Calculate the distance between the two bodies
To find the distance between the two bodies, we need to calculate the absolute difference between their positions:
\[ \text{Distance} = |s_1 - s_2| \]
Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ \text{Distance} = |(v_0 t - \frac{1}{2} g t^2) - (-v_0 t - \frac{1}{2} g t^2)| \]
This simplifies to:
\[ \text{Distance} = |v_0 t - \frac{1}{2} g t^2 + v_0 t + \frac{1}{2} g t^2| \]
Combining like terms gives:
\[ \text{Distance} = |2 v_0 t| \]
Since distance is always positive, we can drop the absolute value:
\[ \text{Distance} = 2 v_0 t \]

### Conclusion
The distance between the two bodies after time \( t \) is:
\[ \text{Distance} = 2 v_0 t \]