Question

A man travelling at \[10.8kmph\] in topless car on a rainy day. He holds an umbrella at an angle of ${37^\circ }$ with the vertical so that he does not get wet. If raindrops fall vertically downwards, what is rain velocity?
A) $1m{s^{ - 1}}$
B) $2m{s^{ - 1}}$
C) $3m{s^{ - 1}}$
D) $4m{s^{ - 1}}$

Answer 1

Hint: This question is totally formula based. First of all we need to convert the speed of the man from$kmph$ to $m{s^{ - 1}}$. After that we need to apply the formula for finding the direction of the umbrella to be held. Since the direction of the umbrella to be held is given and the speed of man is also given so just by putting them in the formula we can find the velocity of the rain.

Complete step by step solution:
The velocity of the man in the question is given as ${v_c} = 10.8kmph$
First of all we need to convert it in $m{s^{ - 1}}$.
Therefore, the velocity of the man will become, ${v_m} = 10.8 \times \dfrac{5}{{18}} = 3m{s^{ - 1}}$
Let us assume the velocity of the rain to be ${v_r}$.
The angle with which the man holds the umbrella is ${37^\circ }$
We know that direction of the resultant can be found by,
$\tan {37^\circ } = \dfrac{{{v_m}}}{{{v_r}}}$……… (i)
Also the value of $\tan {37^\circ }$=$\dfrac{3}{4}$
Now, we need to put the value of $\tan {37^\circ }$in equation (i).
After putting the values, we get,
$\Rightarrow \dfrac{3}{4} = \dfrac{3}{{{v_r}}}$
$ \Rightarrow {v_r} = \dfrac{{3 \times 4}}{3} = 4m{s^{ - 1}}$
Therefore, the required value of velocity of the rain is $4m{s^{ - 1}}$.

Hence, option (D), i.e. $4m{s^{ - 1}}$ is the correct choice of the given question.

Note: The man needs to hold the umbrella in the opposite direction of the resultant to save him from the rain. Also, a resultant is the combination of two or more vectors. The quantities which have both magnitude and direction are known as vectors. The direction of the resultant of the velocity of man and the rain is given by the relation $\tan \theta = \dfrac{{{v_m}}}{{{v_r}}}$.