Physics > Motion in One Dimension > 11.0 Rain problem

  Motion in One Dimension
    1.0 Introduction
    2.0 Kinematic variables
    3.0 Motion in one dimension
    4.0 Derivation of the kinematics equation
    5.0 Vertical motion under gravity
    6.0 Analysis of motion through graph
    7.0 Relative motion
    8.0 Simultaneous motion of two bodies
    9.0 River boat problem
    10.0 Aircraft-wind problem
    11.0 Rain problem

11.1 Solved examples

Question: Rain drops are falling vertically with respect to ground with a speed of $4\ m/s$. If a boy starts running with a speed of $3\ m/s$ in horizontal direction. Find the direction in which he should hold the umbrella.

Solution: Writing all the information as below,

${\overrightarrow v _R} = - 4\widehat j$
${\overrightarrow v _B} = 3\widehat i$

From the concept of relative motion we can write,
$${\overrightarrow v _{RB}} = {\overrightarrow v _R} - {\overrightarrow v _B}$$$${\overrightarrow v _{RB}} = - 4\widehat j - 3\widehat i$$or$${\overrightarrow v _{RB}} = - 3\widehat i - 4\widehat j$$
The man holds the umbrella in the direction of ${\overrightarrow v _{RB}}$.


Therefore,
$$\tan \theta = \frac{3}{4}$$$$\theta = 37^\circ $$
The boy will hold the umbrella making angle of $37^\circ$ with the vertical.


Question: A man is coming down on an incline of angle $30^\circ$. When he walks with a speed of $2\sqrt 3 \,m/s$ he keeps his umbrella vertical to protect himself from the rain. The actual speed of the rain is $5\ m/s$. Find the velocity of the rain.


Solution:


$\left| {{{\overrightarrow v }_R}} \right| = 5\,m/s$
${\overrightarrow v _M} = 2\sqrt 3 \left( {\cos 30^\circ \widehat i - \sin 30^\circ \widehat j} \right)$
${\overrightarrow v _{RM}} = a\widehat j$ (As he keeps his umbrella vertical).

Let the velocity of the rain be,
$${\overrightarrow v _R} = b\widehat i + c\widehat j$$Also, $$\sqrt {{b^2} + {c^2}} = 5 \quad ...(i)$$
From the concept of relative motion we can write,
$${\overrightarrow v _{RM}} = {\overrightarrow v _R} - {\overrightarrow v _M}$$$$a\widehat j = \left( {b\widehat i + c\widehat j} \right) - \left[ {2\sqrt 3 \left( {\cos 30^\circ \widehat i + \sin 30^\circ \widehat j} \right)} \right]$$$$a\widehat j = \left( {b - 3} \right)\widehat i + \left( {c + \sqrt 3 } \right)\widehat j$$So, $$b = 3\quad ...(ii)$$ From equation $(i)$ and $(ii)$ we get,
$$\sqrt {{3^2} + {c^2}} = 5$$$$9 + {c^2} = 25$$$${c^2} = 16$$$$c = \pm 4$$
So, the velocity of the rain is,
$${\overrightarrow v _R} = 3\widehat i \pm 4\widehat j$$

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