Question

A pipe which can rotate in a vertical plane is mounted on a cart. The cart moves uniformly along a horizontal path with a speed v1 = 2m/s. At what angle to the horizontal should the pipe be placed so that the drops of rain falling vertically with a velocity v2 = 6m/s moves parallel to the axis of the pipe without touching it's walls?

Answer 1

Answer:

70.5°

Explanation:

Speed of the cart = v1 = 2m/s. (Given)

Velocity of drops of the rain = v2 = 6m/s.  (Given)

Horizontal component of velocity = v1 = v2 Cos Ф

Thus, substituting the values in the equation -

Ф = cos~-1 (v1/v2)

= cos~-1 (2/6)

= cos~-1 (1/3)

= 70.5°

Thus, the angle to which the horizontal pipe be placed so that the drops of rain falling vertically with a velocity and move parallel to the axis of the pipe without touching it's walls should be 70.5°.

Answer 2

Explanation:

tan. we know when we will swim in the west direction so we will come in starting point by velocity of river

so now we are clear that velocity will be in north West direction

hence

answer is 30 north of west