Question

To a man running up an inclined plane of inclination 37° with horizontal at speed of 5m/s, rain appears to be falling at a speed of 6 m/s vertically downwards. Find the actual magnitude and direction of velocity of rain.

Answer 1

magnitude of the rain is under root 481 by 16 add an angle of Cos inverse 80 /481

Answer 2

Answer:

The speed of rain w.r.t ground is $\sqrt{32} \mathrm{~m} / \mathrm{s}$

The velocity of rain w.r.t ground is$-4 \hat{i}-4 \hat{j}\end{aligned}$

Explanation:

Let the velocity of rain w.r.t ground be

$Vr=a^i+b^j$

First case:

The velocity of man w.r.t ground, $\vec{V}_{m}=-4 \hat{i}-3 \hat{j}$

The velocity of rain w.r.t man,

$\begin{aligned}&\Rightarrow \vec{V}_{r m}=\vec{V}_{r}-\vec{V}_{m}=a \hat{i}+b \hat{j}-(-4 \hat{i}-3 \hat{j}) \\&\Rightarrow \vec{V}_{r m}=(a+4) \hat{i}+(b+3)\hat{j} ---(1)\end{aligned}$

As rain seems to be falling vertically to the man $\theta=-90^{\circ}$ [with x-axis]

$\Rightarrow \tan \left(-90^{\circ}\right) \rightarrow-\infty\\ \left[\tan \theta=\frac{Y \text {-component }}{X \text {-component }}\right.$

From Eq.(1),

$\begin{aligned}&\frac{b+3}{a+4} \rightarrow-\infty \\&\Rightarrow a+4=0\end{aligned}$

Or, $a=-4 \ldots(2)$

Second case:

The velocity of man w.r.t ground,

$\begin{aligned}&\vec{V}_{m}=4 \hat{i}+3 \hat{j} \\&\Rightarrow \vec{V}_{r m}=\vec{V}_{r}-\vec{V}_{m}=a \hat{i}+b \hat{j}-(4 \hat{i}+3 \hat{j}) \\&\Rightarrow \vec{V}_{r m}=(a-4) \hat{i}+(b-3) \hat{j}----(3)\end{aligned}$

Substituting From Eq.(2),

$\Rightarrow \vec{V}_{r m}=-8 \hat{i}+(b-3) \hat{j} \quad \ldots \text { (3) }$

As rain seems to be falling at an angle of $\theta=\tan ^{-1}\left(\frac{7}{8}\right)$ with the horizontal (x-axis)

w.r.t the man

$\begin{aligned}&\tan \theta=\frac{7}{8} \\&\Rightarrow \frac{b-3}{-8}=\frac{7}{8} \\&{\left[\because \tan \theta=\frac{\text { Y-component }}{X \text {-component }}\right.} \\&\Rightarrow b-3=-7 \\&\therefore b=-4\end{aligned}$

Hence, the velocity of rain w.r.t ground is

$\begin{aligned}&\vec{V}_{r}=a \hat{i}+b \hat{j} \\&=-4 \hat{i}-4 \hat{j}\end{aligned}$

Now, its magnitude is

$\begin{aligned}&\left|\vec{V}_{r}\right|=\sqrt{(-4)^{2}+(-4)^{2}} \\&\therefore\left|\vec{V}_{r}\right|=\sqrt{32} \mathrm{~m} / \mathrm{s}\end{aligned}$

The speed of rain w.r.t ground is $\sqrt{32} \mathrm{~m} / \mathrm{s}$