**Finding the Time of Flight**

To solve this problem, we need to find the time it takes for the bomb to hit the target after it is released from the plane. We can use the equation of motion:

`h = ut + (1/2)gt^2`

where:
- h is the height of the plane
- u is the initial velocity of the bomb (which is the same as the velocity of the plane)
- g is the acceleration due to gravity (given as 9.8 m/s^2)
- t is the time of flight

Since the height is given in meters and the velocity is given in kilometers per hour, we need to convert the velocity to meters per second. We know that 1 km = 1000 m and 1 hour = 3600 seconds, so:

`u = 360 km/h * (1000 m/km) / (3600 s/h) = 100 m/s`

Substituting the values into the equation, we have:

`1960 m = 100 m/s * t + (1/2) * 9.8 m/s^2 * t^2`

Simplifying the equation, we get a quadratic equation:

`4.9 t^2 + 100 t - 1960 = 0`

Solving this equation, we find two possible values for t: t = 4 seconds or t = -10 seconds. Since time cannot be negative, we discard the negative solution. Therefore, the time of flight is 4 seconds.

**Finding the Horizontal Distance**

To find the horizontal distance the bomb should be released from the target, we can use the formula:

`d = v * t`

where:
- d is the horizontal distance
- v is the horizontal velocity of the plane (which is the same as the velocity of the bomb)
- t is the time of flight

Since the plane is flying in a horizontal direction, the horizontal velocity is equal to the velocity of the plane, which is given as 360 km/h. Converting this to meters per second, we have:

`v = 360 km/h * (1000 m/km) / (3600 s/h) = 100 m/s`

Substituting the values into the formula, we have:

`d = 100 m/s * 4 s = 400 m`

Therefore, the bomb should be released 400 meters from the target in order to hit it.

7.2 km