Astraight river flows east at a speed of 8 mi/h. a boater starts at the south shore of the river and wants to arrive at a point on the north shore of the river directly opposite the starting point. the motorboat has a speed of 16 mi/h relative to the water. in what direction should the boat be headed?

specifically 30º west of the north

Explanation:

This exercise can be solved using the composition of vectors, let's call the speed of the river that goes east (v₁ = 8 mi / h) to the speed of the boat (v₂ = 16 mi / h) and the desired speed from south to north (v₃ =?).  Look at the attachment to be clear about the vectors

Let's use the Pythagorean triangle

    v₃² = v₁² + v₂²

Let's  calculate v₂

    v₂² = v₃² - v₁²

    v₂ = √ (16² - 8²)

    v₂ = 13.86 mi / h

To find the angle we use trigonometry

    sin θ = v1 / v3

    θ = sin⁻¹ (v1 / v3)

    θ = sin⁻¹ (8/16)

    θ = 30º

The boat must be pointed at 30º with respect to the south-north direction, specifically 30º west of the north

answer: i think he can see forward ten seconds and move forward ten seconds but giorno can undo his actions with ger.

explanation:

small = small current large mag large current