Question

A man starts driving in an open gypsy at 1-0 in rainy weather, Assume that speed of gypsy varies with time is v=kt. At t 1 s. he observes rain is falling vertically to him At r-2s, he finds rain drops hitting him at an angle of 45 with vertical. Assuming velocity of rain to be constant, the angle with vortical at which rain is actually faling stan(a) The value of s​

Answer 1

Given : speed of gypsy varies with time is v=kt.

At t 1 s. he observes rain is falling vertically to him

At t -2s, he finds rain drops hitting him at an angle of 45 with vertical.

Velocity of rain to be constant, the angle with vertical at which rain is actually falling is α

To Find : Value of tanα

Solution:

Velocity of rain  s at angle α  with vertical

Vertical velocity = s cosα

Horizontal velocity = s sinα

v=kt

at  t = 1  

v = k  m/s

he observes rain is falling vertically to him

Hence Rain horizontal velocity and man horizontal velocity are same

k = s sinα

at t = 2

v = 2k m/s

man horizontal velocity wrt to Ran =  2k - s sinα

rain vertical velocity = s cosα

angle = 45°  hence both should be same

=>  2k - s sinα =  s cosα

substitute k=s sinα

=> 2 s sinα - s sinα =  s cosα

=> ssinα =  s cosα

=> tanα  = 1

Hence value of tanα is 1

Learn More:

A standing man observes rain falling vertically with a velocity 20m/s ...

brainly.in/question/11211895

To A Stationery Man Rain Is Falling At An Angle 30° With Speed 6 ...

brainly.in/question/11125164

Answer 2

Step-by-step explanation:

Let the rain velocity

v

R

=

v

1

i

˙

+v

2

j

^

In the first case with respect to man

rain is falling downwards so v

1

=μ

i

^

In the second case with respect to man;

V

M

=2μ

i

^

V

RM

=−u

i

^

+V

2

j

^

at an angle θ

Δ0,

V

2

μ

=tanθ

∴

V

R

=μ

i

^

+μcotθ

j

^

Hope its help..