Afishing barge leaves from a dock and travels upstream (against the current) for 4 hours until it reaches its destination 12 miles away. on the return trip the barge travels the same distance downstream (with the current) in 2 hours. find the speed of the barge in still water.

v(b)   = 4,5  mil/h      speed of the barge in still water

Step-by-step explanation:

d = v*t      barge going upstream       12 miles  and 4 hours trip

                barge returning back         12 miles  and 2 hours trip

let call   v(b) barge velocity   and

v(w)   water velocity

d  =  12 (Mil)  =  4 (h)* [(v(b) - v(w)]

  3   =  v(b)  - v(w)       (1)

d  =  12 (mil)  =  2 (h) * [ (v(b) + v(w)]

        6     =   v(b) + v(w)   (2)

Equations (1)  and (2)  is a two system equation. Solving

      from equation  (1)     v(w)  =  v(b)  - 3

By subtitution in equation (2)

         6   =  v(b)  + v(b)  - 3

         9   =  2v(b)

   v(b)   =  9/2      ⇒     v(b)   = 4,5  mil/h

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°