Question

A man moves on a cycle with a velocity of 4 km/hr. The rain appears to fall on him with a velocity of 3 km/hr vertically. The actual velocity of the rain is

Answer 1

Answer:

I think 7km/h is the right answer

Answer 2

Given:

A man moves on a cycle with a velocity of 4 km/hr. The rain appears to fall on him with a velocity of 3 km/hr vertically.

To find:

Actual Velocity of rain.

Diagram:

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(3,3){\vector(1,0){1}}\put(3,3){\vector(-1,0){1}}\put(3,3){\vector(0,-1){1}}\put(3,3){\vector(1,-1){1}}\put(4.25,3){ \vec{v}_{m} }\put(1.5,3){ \vec{v}_{m} }\put(2,3){\vector(1,-1){1}}\put(3,1.75){ \vec{v}_{rm} }\put(2.5,3.25){4}\put(3.15,2.25){3}\put(2,2){ \vec{v}_{r} }\end{picture}}$

Calculation:

First of all , look at the diagram in order to understand the vector diagram of Velocity of man and Velocity of rain.

$1) \: v_{m} = 4 \: km {hr}^{ - 1}$

$2) \: v_{rm} = 3 \: km {hr}^{ - 1}$

Now , as per Pythagoras theorem;

$\therefore \: {(v_{r})}^{2} = {(v_{m})}^{2} + {(v_{rm})}^{2}$

$= > \: {(v_{r})}^{2} = {(4)}^{2} + {( 3)}^{2}$

$= > \: {(v_{r})}^{2} = 16 + 9$

$= > \: {(v_{r})}^{2} = 25$

$= > \: v_{r} = \sqrt{25}$

$= > \: v_{r} = 5 \: km {hr}^{ - 1}$

So, final answer is:

Velocity of Rain w.r.t Ground = 5 km/hr.