**Introduction:**

To find the initial velocity and maximum height reached by an object thrown vertically above the ground, we can use the equations of motion. In this case, we know that the object returns to its initial position after 4 seconds. We can use this information to determine the initial velocity and maximum height.

**Equations of Motion:**

- The equation for displacement is given by:
- s = ut + (1/2)at²
- Where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration.

- The equation for final velocity is given by:
- v = u + at
- Where v is the final velocity, u is the initial velocity, t is the time, and a is the acceleration.

- The equation for time of flight is given by:
- t = 2u/g
- Where t is the time of flight, u is the initial velocity, and g is the acceleration due to gravity.

**Finding the Initial Velocity:**

We are given that the object returns to its initial position after 4 seconds. Since the object is thrown vertically above the ground, the time of flight is equal to twice the time it takes to reach the maximum height. Therefore, we can write the equation as:

4 = 2u/g

Simplifying the equation, we get:

u = (4g)/2
u = 2g

So, the initial velocity of the object is 2g.

**Finding the Maximum Height:**

To find the maximum height reached by the object, we can use the equation for displacement. At the maximum height, the final velocity is 0 m/s. Therefore, we can write the equation as:

0 = u + at

Substituting the values, we get:

0 = 2g + (-g)t

Simplifying the equation, we get:

t = 2 seconds

Therefore, the time taken to reach the maximum height is 2 seconds.

Now, we can use the equation for displacement to find the maximum height. Substituting the values, we get:

s = ut + (1/2)at²

s = (2g)(2) + (1/2)(-g)(2)²

Simplifying the equation, we get:

s = 4g - 2g
s = 2g

So, the maximum height reached by the object is 2g.

**Conclusion:**

The initial velocity of the object is 2g and the maximum height reached is also 2g.