Two cars A and B move in the same direction along a straight line with speed $$100 km/h$$ and $$80 km/h$$, respectively such that  Car $$A$$ is moving ahead of Car $$B$$. A person in car B throws a stone with a speed $$v$$ so that it hit car $$A$$ with a speed of $$5$$ m/s. The value of  $$v$$ is ________ $$km/h$$ :

Let the speeds with respect to the ground be:

Car A : $$u_A = 100$$ km/h
Car B : $$u_B = 80$$ km/h

A passenger sitting in car B throws the stone straight ahead with speed $$v$$ relative to car B. Hence the speed of the stone with respect to the ground is

$$u_s = u_B + v \quad$$ (because both the car and the throw are in the same direction).

The stone must hit car A with a relative speed (speed of approach) of $$5$$ m/s. Therefore, in ground frame, the relative speed between the stone and car A is

$$u_{\text{rel}} = u_s - u_A = (u_B + v) - u_A.$$

This relative speed should equal $$5$$ m/s. First convert $$5$$ m/s to km/h:

$$5\;\text{m/s} = 5 \times \frac{18}{5} = 18 \;\text{km/h}.$$

Set up the equation

$$u_{\text{rel}} = 18$$
$$\Longrightarrow (u_B + v) - u_A = 18$$
$$\Longrightarrow (80 + v) - 100 = 18$$

Solving for $$v$$:

$$v - 20 = 18 \quad\Longrightarrow\quad v = 38 \;\text{km/h}.$$

Hence the required throwing speed is

Option C which is: 38 km/h.