A submarine dives from the water surface at an angle of 31° below the horizontal, following a straight path 75 m long. How far is the submarine then below the water surface?
A)75 m
B)(75 m)/sin 31°
C)(75 m)sin 31°
D)(75 m)cos 31°
E)none of these answers
75 sin 31
To find how far the submarine is below the water surface, we need to determine the vertical component of the submarine's displacement.
The given information tells us that the submarine travels a distance of 75 meters along a straight path. We are also given an angle of 31° below the horizontal, which represents the angle of the submarine's path in relation to the water surface.
To find the vertical component of the submarine's displacement, we can use trigonometry. Specifically, we can use the sine function, which relates the opposite side of a right triangle to the hypotenuse.
The opposite side of the right triangle represents the vertical component of the submarine's displacement, while the hypotenuse represents the 75-meter straight path traveled by the submarine.
Using the sine function, we have:
sin(31°) = opposite/hypotenuse
sin(31°) = vertical displacement/75 m
To find the vertical displacement, we can rearrange the equation:
vertical displacement = sin(31°) * 75 m
Now we can calculate the vertical displacement:
vertical displacement = sin(31°) * 75 m
vertical displacement ≈ 0.515 * 75 m
vertical displacement ≈ 38.625 m
Therefore, the submarine is approximately 38.625 meters below the water surface.
The correct answer is E) none of these answers since none of the given answer choices match the calculated value.