Submarine is traveling parallel to the surface of the water 492 feet below the surface. The submarine begins a constant I sent to the surface so that it will emerge on the surface after traveling 3974 feet from the point of its initial ascent? What horizontal distance did the submarine travel during its dissent?
To find the horizontal distance the submarine traveled during its descent, we need to consider the fact that the submarine is traveling parallel to the surface of the water.
Let's assume that the submarine traveled x feet horizontally during its descent.
Since the submarine is traveling at a constant incline, we can form a right triangle with the horizontal distance it traveled, the vertical distance it ascended, and the hypotenuse representing the direct path from its starting point to where it will emerge on the surface.
Using the Pythagorean theorem, we have:
x^2 + 492^2 = 3974^2
We can solve this equation to find the value of x.
492^2 = 3974^2 - x^2
242,064 = 15,797,876 - x^2
x^2 = 15,555,812
x = √15,555,812
x ≈ 3,946.2 feet
Therefore, the submarine traveled approximately 3,946.2 feet horizontally during its descent.
√(3974^2-492^2) ≈ 3943 ft
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