Rain Drop

tan 45 = velocty of rain / velosty of car

$tan { 45\quad =\quad \frac { Vrain }{ Vcar } } \\ 1\quad =\quad \frac { Vrain }{ 3 } \\ Vrain\quad =\quad 3\quad ㎧$

tan(45)=v/3 so v = 3*tan(45)=3 m/s

tan45-= 1= 3/x therefore x= 3m/s

An observer in the reference frame of the window, which is moving at a constant speed of 3 m / s see the raindrops falling in a direction that is 45 ° to the vertical (or horizontal). So the sum vector car speeds, and rain should form a rectangle and an isosceles triangle, where the speed vectors are the legs. Therefore, for a reference outside the car the rain must fall vertically with a speed of module 3 m / s...

Um observador no referencial da janela, que está se movendo a uma velocidade constante de 3 m/s vê as gotas de chuva caindo em uma direção que faz 45° com a vertical (ou horizontal). Logo a soma das velocidades vetoriais do carro e da chuva devem formar um triângulo retângulo e isósceles, onde os vetores velocidades são os catetos. Logo, para um referencial fora do carro a chuva deve cair verticalmente com uma velocidade de módulo 3 m/s.

this question is wrong.because question says that rain fall on car is 45 angle but in bracket given that rain fall straight down. secondly question asked that what is speed of rain but in bracket given that assume speed of rain is constant.how this possible???????

tan 45 = |velocity of car|/|velocity of rain| ... so answer is 3m/s ( ' | | ' means the mod value of the velocity)

3

TAN45=V/3 (TAN45=1) 1=V/3 V=3

3m/s

the answer is like this : tan(45) = 1 then. tan(45) *3=3m/s

tan45=vel of rain/vel of car

:)

tan45=velocity of car/velocity of rain drop (velocity of car is 3m/s and rain is x )so tan 45=1,1=3/x ,=3m/s

v=u+at; v=0,u=30; then t=3.06 tatal time required T=6.12

IT shuold b 3 m/s

It's ok if , tan 45 degree = Velocity of rain / Velocity of car = 1 ; The velocity of rain = The velocity of car = 3 m/s.

what about if , There is a Right triangle whose angles are : 90 degree , 45 degree & 45 degree . so Opposite side ( the velocity of rain) = Adjacent side ( the velocity of the car ) = 3 !

the ans is 3 m/s

Direction of rain with respect to car is given by : tan45= Vel. of Rain/ Vel. of Car 1=x/3 (say 'x' be velocity of rain) Therefore, x=3

tan 45 = Velocity of rain/Velocity of car

if v is vel of rain, then tan45 =v/3, or v = 3 * tan45 = 3 * 1 =3.

since its clearly mentioned to assume the rain drops fall straight down thus we take tan angle thus tan45=velocity of rain drops/3 = 3m/s

tan 45=velocityof drop downward/velocity of car....... tan45=x/3...... x=3

tan45=Vrain/Vcar => 1=Vrain/3 =>1*3=Vrain, or Vrain=3 m/s

tan(45)=v/3 so v = 3*tan(45)=3 m/s

tan45=x/3 1=x/3 3=x

tan45= V rain/ Vcar

By Right Angled Triangle, tan = OPP Side/ADJ Side i.e., tan 45 = rain drop speed/ car speed 1 = x/3 from these x=3

tan45= velocityof rain(Vr)/ velocity of car(Vc) 1=Vr/Vc 1=Vr/3 Vr=3

The above problem has solution = 3 m/s

tan 45 = (velocty of rain) / (velosty of car)

Right 45-45-90 vector triangles, anyone?

The window is oblique not the rain ever ..

Since rain falling straight will make a issocelous triangle having 45 degree angle. iIt has 90 degree anle on screen of car. speed of car is 3 miles/sec , so base of triangle is 3,and hight of rtriangle will be 3 ( isocellous trianle).therefore,Answer is 3

If speed of rain is r then tan 45=3/r => r = 3 m/s

Isn't it supposed to be $3 \sqrt{2}$ ?