A kayaker needs to paddle north across a 100 -m wideharbor. The tide is going out, creating a tidal current that flowsto the east at 2.0 m/s. The Kayaker can paddle with a speed of3.0m/s.
a. In which direction shoule he paddle in order to travel straightacross the harbor ?
b. How long will it take him to cross ?

A body diagram and step by step precise solution willgive me a better understanding .. thanks for you help..

a) v_Nort = 2.236 m / s

, θ = 56.3º, b)  t=  53.76 s

Explanation:

This exercise should use the addition of vectors, we have a kayak speed of v₁ = 3 m / s and an eastward speed of water v₂ = 2.0 m / s.

They ask us to cross the river that is to the north, we see that the speed of the kayak is the hypotenuse of the triangula, see attached

              v₁² = v_nort² + v₂²

              v _nort² = v₁² –v₂²

              v_nort = √ (3² - 2²)

              v_Nort = 2.236 m / s

For the angle we can use trigonometry

              tan θ = v₂ / v₁

              θ = tan⁻¹ v₂ / v₁

              θ = tan⁻¹ 2/3

              θ = 33.7º

This angle measured from the positive side of the x axis is

              θ = 90 - 33.7

              θ = 56.3º

b)  we look for the northward component of this speed

          sin 56.3 = / v_nort

          v_{y} = v_nort sin 56.3

          v_{y} = 2.236 sin 56.3

          v_{y} =  1.86 m/s

The time is

          v_{y} = y/t

          t = y/v_{y}

          t =100/ 1.86

          t=  53.76 s

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas