Question

When a man moves down the inclined plane with a constant speed with a $5\;{\text{m}}{{\text{s}}^{ - 1}}$ which makes an angle of $37^\circ $ with the horizontal, he finds that the rain is falling vertically downward. When he moves same up the same inclined plane with the same speed, he finds that the rain makes an angle $\theta = {\tan ^{ - 1}}\left( {\dfrac{7}{8}} \right)$ with the horizontal. The speed of the rain is:
A) $\sqrt {116} \;{\text{m}}{{\text{s}}^{ - 1}}$
B) $\sqrt {32} \;{\text{m}}{{\text{s}}^{ - 1}}$
C) $5\;{\text{m}}{{\text{s}}^{ - 1}}$
D) $\sqrt {73} \;{\text{m}}{{\text{s}}^{ - 1}}$

Answer 1

Hint: Here we use the concept of the relative motion. In the relative motion source moves relative to the observer or observer relative to the source.

Complete step by step answer:
Given: The angle of the incline is $\theta = 37^\circ $, the speed of the man is $v = 5\;{\text{m}}{{\text{s}}^{ - 1}}$, the angle of the ran for up to incline is $\alpha = {\tan ^{ - 1}}\left( {\dfrac{7}{8}} \right)$.
The velocity of man is given as,
$\Rightarrow$ ${V_m} = - \left( {v\cos \theta } \right)\hat i - \left( {v\sin \theta } \right)\hat j......\left( 1 \right)$
The velocity of the rain is given as,
$\Rightarrow$ ${v_r} = {v_x}\hat i - {v_y}\hat j.....\left( 2 \right)$
Here, ${v_x}$ and ${v_y}$ are components of the velocity of the rain along the x and y axis.
The velocity of the rain relative to man is given as,
$\Rightarrow$ ${v_{rm}} = {v_r} - {v_m}......\left( 3 \right)$
Rearrange the equation (1), equation (2) and equation (3).
$\Rightarrow$ ${v_{mr}} = \left[ {{v_x} + v\cos \theta } \right]\hat i + \left[ {{v_y} - v\sin \theta } \right]\hat j$
Substitute $v = 5\;{\text{m}}{{\text{s}}^{ - 1}}$ and $\theta = 37^\circ $in the equation (1) to find the relative velocity of the rain.
$\Rightarrow$ ${v_{rm}} = \left[ {{v_x} + \left( {5\;{\text{m}}{{\text{s}}^{ - 1}}} \right)\left( {\cos 37^\circ } \right)} \right]\hat i + \left[ {{v_y} - \left( {5\;{\text{m}}{{\text{s}}^{ - 1}}} \right)\left( {\sin 37^\circ } \right)} \right]\hat j$
$\Rightarrow$ ${v_{rm}} = \left( {{v_x} + 4\,{\text{m}}{{\text{s}}^{ - 1}}} \right)\hat i + \left( {{v_y} - 5\,{\text{m}}{{\text{s}}^{ - 1}}} \right)\hat j......\left( 4 \right)$
The rain is falling vertically downward while man moves down the incline plane, so the x component of the velocity of rain relative to man becomes zero.
Equate the x component of velocity in equation (4) to zero to find the x component of the velocity of the rain.
$\Rightarrow$ ${v_x} + 4\,{\text{m}}{{\text{s}}^{ - 1}} = 0$
$\Rightarrow$ ${v_x} = - 4\,{\text{m}}{{\text{s}}^{ - 1}}$
When the man is moving up the incline then the velocity of man is given as,
$\Rightarrow$ ${v_M} = \left( {v\cos \theta } \right)\hat i + \left( {v\sin \theta } \right)\hat j......\left( 5 \right)$
Rearrange the equation (2) and equation (5) to find the relative velocity of rain when man is moving up.
$\Rightarrow$ ${v_{rM}} = \left[ {{v_x} - v\cos \theta } \right]\hat i + \left[ { - {v_y} - v\sin \theta } \right]\hat j......\left( 6 \right)$
Substitute $v = 5\;{\text{m}}{{\text{s}}^{ - 1}}$ , ${v_x} = - 4\,{\text{m}}{{\text{s}}^{ - 1}}$and $\theta = 37^\circ $in the equation (6) to find the y component of the velocity of the rain.
$\Rightarrow$ ${v_{rM}} = \left[ { - 4\,{\text{m}}{{\text{s}}^{ - 1}} - \left( {5\;{\text{m}}{{\text{s}}^{ - 1}}} \right)\left( {\cos 37^\circ } \right)} \right]\hat i + \left[ {{v_y} - \left( {5\;{\text{m}}{{\text{s}}^{ - 1}}} \right)\left( {\sin 37^\circ } \right)} \right]\hat j$
$\Rightarrow$ ${v_{rM}} = \left( { - 8\,{\text{m}}{{\text{s}}^{ - 1}}} \right)\hat i + \left[ { - {v_y} - 3\,{\text{m}}{{\text{s}}^{ - 1}}} \right]\hat j......\left( 7 \right)$
The angle of the rain appears $\alpha = {\tan ^{ - 1}}\left( {\dfrac{7}{8}} \right)$ to man when man moves up to the plane.
Equate the y component of the relative velocity of the rain when man moves up to plane to y component of the appeared angle.
$\Rightarrow$ $ - {v_y} - 3\,{\text{m}}{{\text{s}}^{ - 1}} = - 7\;{\text{m}}{{\text{s}}^{ - 1}}$
$\Rightarrow$ ${v_y} = 4\;{\text{m}}{{\text{s}}^{ - 1}}$
Use the expression of resultant to find the speed of the rain is,
$\Rightarrow$ $V = \sqrt {v_x^2 + v_y^2} .........\left( 8 \right)$
Substitute ${v_x} = - 4\,{\text{m}}{{\text{s}}^{ - 1}}$ and ${v_y} = 4\;{\text{m}}{{\text{s}}^{ - 1}}$ in the equation (8) to find the velocity of rain.
$\Rightarrow$ $V = \sqrt {{{\left( { - 4\,{\text{m}}{{\text{s}}^{ - 1}}} \right)}^2} + \left( {4\,{\text{m}}{{\text{s}}^{ - 1}}} \right)} $
$\Rightarrow$ $V = \sqrt {32} \,{\text{m}}{{\text{s}}^{ - 1}}$

Therefore, the speed of the rain is $\sqrt {32} \,{\text{m}}{{\text{s}}^{ - 1}}$ and the option (B) is correct.

Note: Be careful to calculate the x and y components of the velocity of the rain and relative speed of the rain. Relative velocity is frame independent because it is the relative change.