Two bodies of same mass tied with an inelastic string of length `l` lie together. One of them is projected vertically upwards with velocity `sqrt(6gl)`. Find the maximum height up to which the centre of mass system of the two masses rises.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the System
We have two bodies of the same mass \( m \) connected by an inelastic string of length \( l \). One body (let's call it A) is projected upwards with an initial velocity of \( v_0 = \sqrt{6gl} \), while the other body (B) remains stationary initially.

### Step 2: Determine the Initial Conditions
The initial position of the center of mass (CM) of the system can be calculated as:
\[ h_{\text{CM, initial}} = \frac{h_A + h_B}{2} = \frac{0 + \frac{l}{2}}{2} = \frac{l}{4} \]
where \( h_A = 0 \) (the height of A) and \( h_B = \frac{l}{2} \) (the height of B).

### Step 3: Analyze the Motion of Body A
As body A moves upwards, it will continue to rise until the string becomes taut. We need to find the velocity of body B when the string becomes taut.

### Step 4: Apply Conservation of Momentum
When the string becomes taut, both bodies will have the same speed \( v' \). By conservation of momentum:
\[ m v_0 + m \cdot 0 = m v' + m v' \]
This simplifies to:
\[ v_0 = 2v' \]
Thus, we find:
\[ v' = \frac{v_0}{2} = \frac{\sqrt{6gl}}{2} = \frac{\sqrt{6}}{2} \sqrt{gl} \]

### Step 5: Calculate the Maximum Height of the Center of Mass
The center of mass will rise with the velocity \( v' \) until both bodies reach their maximum height. The maximum height can be calculated using the kinematic equation:
\[ v^2 = u^2 + 2as \]
where \( v = 0 \) (final velocity at maximum height), \( u = v' \), and \( a = -g \) (acceleration due to gravity).

Setting up the equation:
\[ 0 = \left(\frac{\sqrt{6}}{2} \sqrt{gl}\right)^2 - 2g h \]
This gives:
\[ 0 = \frac{6}{4} gl - 2gh \]
Rearranging gives:
\[ 2gh = \frac{6}{4} gl \]
\[ h = \frac{6l}{8} = \frac{3l}{4} \]

### Step 6: Calculate the Final Height of the Center of Mass
The final height of the center of mass is the initial height of the center of mass plus the height it rises:
\[ h_{\text{CM, final}} = h_{\text{CM, initial}} + h = \frac{l}{4} + \frac{3l}{4} = l \]

### Final Answer
The maximum height up to which the center of mass of the system rises is:
\[ \boxed{l} \]

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