Using relative motion (2D) to get rain speed & direction when running

[aaronstonedd]

Homework Statement

A person moving towards east with a speed 'v' observes the rain to be falling vertically downwards. When he doubles his speed, the rain appears to come at 30° angle with the vertical. Find the actual speed and direction of rain with vertical.

Homework Equations

Velocity of A with respect to B = Velocity of A – Velocity of B
(V_AB = V_A – V_B)

The Attempt at a Solution

Let the actual speed of rain be V_R. Let the person's speed be represented by V_P.
Using \widehat{i}-\widehat{j}-\widehat{k} notation for vectors, V_P = v \widehat{i}

V_R = V_Rsinθ \widehat{i} – V_Rcosθ \widehat{j}

V_RP = V_R – V_P
V_RP = V_Rsinθ \widehat{i} – V_Rcosθ \widehat{j} – v \widehat{i} = (V_Rsinθ – v) \widehat{i} –V_Rcosθ \widehat{j}

∵ Rain is vertically downwards, ∴ (V_Rsinθ – v) \widehat{i} = 0

So, V_Rsinθ = v

Now, let V_P be 2v \widehat{i}.

∴ V_RP = V_R – V_P
= +V_Rsinθ \widehat{i} –V_Rcosθ \widehat{j} –2v \widehat{i}
= (+V_Rsinθ –2v) \widehat{i} –V_Rcosθ \widehat{j}
= –v \widehat{i} –vtanθ \widehat{j}

Now \frac{-v}{-vtan\theta} = tan30° = √3̅

∴ tanθ = \frac{1}{\sqrt{3}}

∴ θ = 60°.

V_R = \frac{2v}{\sqrt{3}}.


Both these answers are wrong, and I don't know how or why. I'm in Class 11. The correct answers should be θ = 30° and ∴ V_R = 2v. What am I missing here?

 

[mfb]

tan 30° is not the square root of 3.

It is confusing that you use the same symbol for vectors and scalar quantities.

 

[aaronstonedd]

Two mistakes

Sorry, I made another mistake. Though you're right, sin30° = \frac{1}{\sqrt{3}}, I made another mistake: -V_Rcosθ \widehat{j} ≠ -vtanθ \widehat{j}, but, = -vcotθ \widehat{j}.

So still, θ = 60°. Why is it so? (I know my answer's wrong, but I don't know why.)