Question

A man moves on a horizontal road towards east at
a speed of 1 km/hr and the rain appears to him
vertical at a speed of 2 km/hr. The actual speed of
the rain (in km/hr) is​

Answer 1

Given:

$\sf{V_{m}}$ be the speed of man = 1 km/hr

$\sf{V_{r}}$ be the speed of rain = 2 km/hr

Note: It is given that, the rain falls vertically.

So:

We will use:

$\boxed{\sf{V_{rm}= \sqrt{( {vm})^{2} +( {vr})^{2} }}}$

Now:

$\sf{V_{rm}= \sqrt{( {vm})^{2} +( {vr})^{2} }}$

$\sf{V_{rm} = \sqrt{(1)^{2} + ( {2})^{2} }}$

$\sf{V_{rm} = \sqrt{1 + 4}}$

$\sf{V_{rm} = \sqrt{5}}$

$\sf{V_{rm} = 2.236 \: km/hr}$

Therefore:

The actual speed of the rain is 2.236 km/hr.

Answer 2

Explanation:

Let Vrm = 2

Vr = √(4+1)

= √5 = 2.236 km/h