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**Hint:**If two bodies are moving along the same line in the same direction with velocities ${V_A}$ and ${V_B}$ relative to earth then the magnitude of the velocity of \[A\] relative to \[B\] is given by ${V_A} - {V_B} = {V_{AB}}$. The velocity of one object with respect to some frame is known as the reference frame. Draw a free-body using the data given above and determine the velocity of the rain with respect to the ground.

**Complete step by step solution:**

The velocity of one object with respect to some frame is known as the reference frame. Objects move in a straight line in one-dimension motion. So, there are only two possible cases are objects are moving in the same direction and Objects are moving in the opposite direction

The velocity of man =$\vec u = u\hat i$

Let the velocity of the rain =$\vec v = x\hat i + y\hat j$

Let us find the velocity of rain with respect to man=${V_R} = \left( {x - u} \right)\hat i + y\hat j$

Rain is falling vertically down then the \[\;x\] component is zero and then the \[y\] component is along the negative direction.

Hence, the velocity of rain=$u\hat i + y\hat j$

When he doubles the speed velocity of rain with respect to man =${V_R}$

${V_R} = u\hat i + y\hat j - 2u\hat i$

${V_R} = - u\hat i + y\hat j$

Now the rain is falling at the angle $\theta $ to the vertical

$\tan \theta = \dfrac{{ - u}}{y}$

The velocity of rain with respect to ground

$V = x\hat i + y\hat j$

$V = u\hat i - \dfrac{y}{{\tan \theta }}\hat j$

**Hence option $\left( B \right)$ is the right option.**

**Note:**Objects move in a straight line in one-dimension motion. So, there are only two possible cases: objects are moving in the same direction and Objects are moving in the opposite direction. If two bodies are moving along the same line in the same direction with velocities ${V_A}$ and ${V_B}$ relative to earth then the magnitude of the velocity of \[A\] relative to B is given by ${V_A} - {V_B} = {V_{AB}}$.

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