Question

For a man walking towards east with a velocity of 5 km/hr rain appears to fall vertically down. When the man begins to walk towards west with a velocity of 3 km/hr the rain appears to fall at an angle of 60° to the vertical. The magnitude of velocity of the rain when the man stops walking is √139/ N kmph. Then N is equal to​

Answer 1

Answer:

The value of N = 3.

Explanation:

Given when a man is walking towards east with a velocity = 5 km/hr

If we take east along x direction, then $\vec v_m=5\hat i~ km/hr$

When he is walking east, rain appears to fall vertically down.

Let us resolve the velocity vector of rain into two components, $\vec v_r=v_x\hat i+v_y\hat j=v_x\hat i-v_y\hat j$

since rain is falling downwards, y is taken negative.

Then the relative velocity in this case can be $\vec v_{r/m}= v_x\hat i+v_y\hat j-5\hat i=(v_x-5)\hat i-v_y\hat j$

Along x-direction, $\vec v_{r/m}= 0\implies v_x-5=0$

Therefore,  $v_x=5 km/hr$

Given when a man is walking towards west with a velocity = 3 km/hr

Since he west, towards -x direction, then $\vec v_m=-3\hat i km/hr$

Then the relative velocity in this case can be $\vec v_{r/m}= v_x\hat i-v_y\hat j-(-3)\hat i$

$=(v_x+3)\hat i-v_y\hat j=(5+3)\hat i-v_y\hat j=8\hat i-v_y\hat j$

When he is walking west, rain appears to fall at 60°.

Therefore, $tan 60^o=\dfrac{v_x}{v_y} =\dfrac{8}{v_y}$        

$\implies \sqrt{3} =\dfrac{8}{v_y}\implies v_y=\dfrac{8}{\sqrt{3} }$  

Thus the resultant velocity of rain is,

$|\vec v_r|=\sqrt{v_x^{2} +v_y^{2}} =\sqrt{5^{2} +\Bigg(\dfrac{8}{\sqrt{3} }\Bigg)^{2}}$

$=\sqrt{25+\dfrac{64}{3}} = \sqrt\dfrac{75+64}{3}}=\sqrt\dfrac{139}{3}} kmph$

Given, the magnitude of velocity of the rain when the man stops walking is $\sqrt{\dfrac{139}{N}} kmph$.

Comparing this with the obtained result, value of N = 3.

Hence, the value of N = 3.

Answer 2

Answer:

The value of $\mathrm{N}=3 .$

Explanation:

Given when a man is walking towards east with a velocity $=5 \mathrm{~km} / \mathrm{hr}$

If we take east along $x$ direction, then $\vec{v}_{m}=5 \hat{i} \mathrm{~km} / \mathrm{hr}$

When he is walking east, rain appears to fall vertically down.

Let us resolve the velocity vector of rain into two components,

$\vec{v}_{r}=v_{x} \hat{i}+v_{y} \hat{j}=v_{x} \hat{i}-v_{y} \hat{j}$

since rain is falling downwards, $y$ is taken negative.

Then the relative velocity in this case can be

$\vec{v}_{r / m}=v_{x} \hat{i}+v_{y} \hat{j}-5 \hat{i}=\left(v_{x}-5\right) \hat{i}-v_{y} \hat{j}$

Along $x$-direction, $\vec{v}_{r / m}=0 \Longrightarrow v_{x}-5=0$

Therefore, $v_{x}=5 \mathrm{~km} / \mathrm{hr}$

Given when a man is walking towards west with a velocity $=3 \mathrm{~km} / \mathrm{hr}$

Since he west, towards$-x$ direction, then $\vec{v}_{m}=-3 \hat{i} k m / h r$

Then the relative velocity in this case can be $\vec{v}_{r / m}=v_{x} \hat{i}-v_{y} \hat{j}-(-3) \hat{i} =\left(v_{x}+3\right) \hat{i}-v_{y} \hat{j}=(5+3) \hat{i}-v_{y} \hat{j}=8 \hat{i}-v_{y} \hat{j}$

When he is walking west, rain appears to fall at $60^{\circ} .$

Therefore, $\tan 60^{\circ}=\frac{v_{x}}{v_{y}}=\frac{8}{v_{y}}$

$\Longrightarrow \sqrt{3}=\frac{8}{v_{y}} \Longrightarrow v_{y}=\frac{8}{\sqrt{3}}$

Thus the resultant velocity of rain is,

$\begin{aligned}&\left|\vec{v}_{r}\right|=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{5^{2}+\left(\frac{8}{\sqrt{3}}\right)^{2}} \\&=\sqrt{25+\frac{64}{3}}=\sqrt{\frac{75+64}{3}}=\sqrt{\frac{139}{3}} \mathrm{kmph} \\&\text { Given, the magnitude of velocity of the rain when the man stops walking is } \\&\sqrt{\frac{139}{N}} \mathrm{kmph}\end{aligned}$

Comparing this with the obtained result, value of N = 3.

Hence, the value of N = 3.