Two bodies are projected vertically upwards from one point with the same initial velocity `v_0.` The second body is projected `t_0 s` after the first. How long after will the bodies meet?

Explanation

To solve the problem of when the two bodies meet after being projected vertically upwards, we can follow these steps:

### Step 1: Understand the motion of both bodies
Let:
- Body 1 is projected at time \( t = 0 \) with initial velocity \( v_0 \).
- Body 2 is projected at time \( t = t_0 \) with the same initial velocity \( v_0 \).

### Step 2: Write the displacement equations
Using the equation of motion for both bodies, we can express their displacements:

For Body 1 (projected at \( t = 0 \)):
\[ s_1 = v_0 t - \frac{1}{2} g t^2 \]

For Body 2 (projected at \( t = t_0 \)):
Since Body 2 is projected \( t_0 \) seconds after Body 1, its time of flight when the collision occurs will be \( t - t_0 \):
\[ s_2 = v_0 (t - t_0) - \frac{1}{2} g (t - t_0)^2 \]

### Step 3: Set the displacements equal
At the time of collision, the displacements of both bodies will be equal:
\[ s_1 = s_2 \]
Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ v_0 t - \frac{1}{2} g t^2 = v_0 (t - t_0) - \frac{1}{2} g (t - t_0)^2 \]

### Step 4: Simplify the equation
Expanding the right side:
\[ v_0 t - \frac{1}{2} g t^2 = v_0 t - v_0 t_0 - \frac{1}{2} g (t^2 - 2t t_0 + t_0^2) \]
Now, simplifying both sides:
\[ v_0 t - \frac{1}{2} g t^2 = v_0 t - v_0 t_0 - \frac{1}{2} g t^2 + g t t_0 - \frac{1}{2} g t_0^2 \]

### Step 5: Cancel common terms
Cancelling \( v_0 t \) and \( -\frac{1}{2} g t^2 \) from both sides:
\[ 0 = -v_0 t_0 + g t t_0 - \frac{1}{2} g t_0^2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ g t t_0 = v_0 t_0 + \frac{1}{2} g t_0^2 \]
Dividing both sides by \( t_0 \) (assuming \( t_0 \neq 0 \)):
\[ g t = v_0 + \frac{1}{2} g t_0 \]

### Step 7: Solve for \( t \)
Now, solving for \( t \):
\[ t = \frac{v_0}{g} + \frac{1}{2} t_0 \]

### Final Answer
Thus, the time after which the two bodies will meet is:
\[ t = \frac{v_0}{g} + \frac{1}{2} t_0 \]

---