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**Hint:**This question requires application of conservation of momentum theorem. Ground is obviously at rest, so velocity of the ground is 0.

**Complete step by step answer:**

Given:

Mass of the trolley \[ = 40kg\] (\[{m_1}\])

Mass of the man \[ = 80kg\](\[{m_2}\])

Speed of the trolley \[ = 2m/s\]

Initially, both man and trolley move together as a system, therefore, we can say the system moves with velocity\[ = 2m/s\].

Applying conservation of momentum theorem, which states that:

The theorem states that momentum can never be created nor be destroyed but can be changed through action of forces governed by Newton’s laws of motion.

Initial momentum and final momentum remains the same.

After the man jumps off the trolley, the man acquires a different velocity, and the cart continues to move with the same velocity.

Conservation of momentum is just an application of Newton’s Third law of motion.

Let us assume the following:

\[{v_1}\] Be the initial velocity of the cart along with the man.

\[{v_2}\] Be the Final velocity of the cart after the man has jumped off.

Applying conservation of momentum theorem:

\[({m_{1 + }}{m_{_2}}){v_1} = {m_2}{v_2}\]

Putting the values:

\[(80 + 40)2 = 40 \times {v_2}\]

Solving the equation:

\[{v_2} = 1.5m/s\]

Therefore, velocity becomes \[1.5m/s\].

**Therefore, velocity with respect to ground is \[ = 0 - 1.5m/s\] = $1.5 m/s$.**

**Note:**Relative velocity is defined as the velocity of object A in the rest frame of another object B. Inertial frame of an object is either at rest or moves with a constant velocity. In case of collisions, the law of momentum can also be applied. Collisions can be of two types:

Elastic and inelastic collisions.

Momentum defines, the quantity of motion that an object has, Mathematically expressed as:

\[p = mv\]

Where:

\[p\] is the momentum of the object.

\[m\] is the mass of the object

\[v\] is the velocity of the object.

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