Question

A man is standing on a road and observes that rain is falling at angle ${45}^{\circ }$ with the vertical. The man starts running on the road with constant acceleration $0.5{\text{m/s}}^2$. After a certain time from the start of the motion, it appears to him that rain is still falling at angle ${45}^{\circ }$ with the vertical, with speed $2\sqrt{2}\text{m/s}$. Motion of the man is in the same vertical plane in which the rain is falling. Then which of the following statement(s) are true

• A. It is not possible
• B. Speed of the rain relative to the ground is $2\text{m/s}$
• C. Speed of the man when he finds rain to be falling at an angle of ${45}^{\circ }$ with the vertical is $4\text{m/s}$.
• D. The man has travelled a distance of $16\text{m}$ on the road by the time he finds rain to be falling at angle ${45}^{\circ }$.

Answer 1

Answer: (C), (D)

The correct options are
C Speed of the man when he finds rain to be falling at an angle of ${45}^{\circ }$ with the vertical is $4\text{m/s}$.
D The man has travelled a distance of $16\text{m}$ on the road by the time he finds rain to be falling at angle ${45}^{\circ }$.

${\overrightarrow{V}}_{rg}={\overrightarrow{V}}_{rm}+{\overrightarrow{V}}_{mg}$

As the man speeds up, the angle with the vertical keeps on decreasing and the rain becomes vertical w.r.t. to the man at some point in time. On increasing the speed further, the angle with the vertical increases on the other side as shown is figure.

Taking positive x-axis to be rightwards and positive y-axis to be upwards.
${\overrightarrow{V}}_{rm}={\overrightarrow{V}}_{rg}-{\overrightarrow{V}}_{mg}$
$V_{rm}\cos {45}^{\circ }=V_{rg}\cos {45}^{\circ }$
$V_{rm}=2\sqrt{2}\text{m/s}=V_{rg}$
$V_{rm}\sin {45}^{\circ }=V_{mg}-V_{rg}\cos {45}^{\circ }$
$V_{mg}=2\sqrt{2}\frac{1}{\sqrt{2}}+2\sqrt{2}\frac{1}{\sqrt{2}}=4\text{m/s}$

Using $v^2=u^2+2as$ for the motion of man,
$v^2=u^2+2as$
or $4^2=0+2\times 0.5s$
or $s=16\text{m}$