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Question: A man running along a straight road with uniform velocity \(\vec u = u\hat i\) feels that the rain i...

A man running along a straight road with uniform velocity u⃗=ui^\vec u = u\hat i feels that the rain is falling vertically down along −j-j. If he doubles his speed, he finds that the rain is coming at an angle θ\theta with the vertical. The velocity of the rain with respect to the ground is:
(A)ui−utan⁡θj^\left( A \right)ui - u\tan \theta \hat j
(B)ui−utan⁡θj^\left( B \right)ui - \dfrac{u}{{\tan \theta }}\hat j
(C)utan⁡θ−uj^\left( C \right)u\tan \theta - u\hat j
(D)utan⁡θi−uj^\left( D \right)\dfrac{u}{{\tan \theta }}i - u\hat j

Explanation

Solution

If two bodies are moving along the same line in the same direction with velocities VA{V_A} and VB{V_B} relative to earth then the magnitude of the velocity of AA relative to BB is given by VA−VB=VAB{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 =u⃗=ui^\vec u = u\hat i
Let the velocity of the rain =v⃗=xi^+yj^\vec v = x\hat i + y\hat j

Let us find the velocity of rain with respect to man = VR=(x−u)i^+yj^{V_R} = \left( {x - u} \right)\hat i + y\hat j
Rain is falling vertically down then the   x\;x component is zero and then the yy component is along the negative direction.
Hence, the velocity of rain=ui^+yj^u\hat i + y\hat j
When he doubles the speed velocity of rain with respect to man =VR{V_R}
VR=ui^+yj^−2ui^{V_R} = u\hat i + y\hat j - 2u\hat i
VR=−ui^+yj^{V_R} = - u\hat i + y\hat j
Now the rain is falling at the angle θ\theta to the vertical
tan⁡θ=−uy\tan \theta = \dfrac{{ - u}}{y}
The velocity of rain with respect to ground
V=xi^+yj^V = x\hat i + y\hat j
V=ui^−ytan⁡θj^V = u\hat i - \dfrac{y}{{\tan \theta }}\hat j
Hence option (B)\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 VA{V_A} and VB{V_B} relative to earth then the magnitude of the velocity of AA relative to B is given by VA−VB=VAB{V_A} - {V_B} = {V_{AB}}.