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physics

A river flows due east at 1.50 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 9.2 m/s due north relative to the water.
(a) What is the velocity of the boat as viewed by an observer on the shore?
m/s at ° east of north
(b) If the river is 301 m wide, how far downstream is the boat when it reaches the north shore?
m

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Anonymous

  1. X = 1.50 m/s.
    Y = 9.2 m/s.

    a. tanA = Y/X = 9.2/1.5 = 6.13333.
    A = 80.7o, CCW.
    A = 90-80.7 = 9.26o East of North.

    b. tanA = 301/d.
    d = 301/tanA = 301/tan80.7 = 49.3 m.

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    Henry

  2. a, V^2 = X^2 + Y^2 = 86.89
    V = 9.32 m/s.

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    Henry

  4. my mans henry has the most confusing explanations

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    no

  5. I'm a physics student getting cucked by Henry on the second part of A. Here is actually how to do it along with the other stuff

    a)1.5^2+9.2^2=86.89 Square 86.89=9.3

    a2) 9.2/1.5=6.133333 at this point use the inverse of tan to get 80.7, then do 90-80.7. to get 9.3.

    b)301m/9.2 m/s due north= 32.72*1.5 m/s=49.08

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    AwesomeMage24

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