# A child rides a toboggan down a hill that descends at an angle of 27.2 to the horizontal. The hill is 20.1 m long . What is the horizontal component of the child's displacement? Answer in units...

A child rides a toboggan down a hill that descends at an angle of 27.2 to the horizontal. The hill is 20.1 m long .

What is the horizontal component of the child's displacement?

Answer in units of m.

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The sign (positive or negative) of the horizontal component of a child's displacement would depend on the choice of the direction of the horizontal (x) axis. I am going to direct the x-axis in the same direction as the child is sliding down the hill. Then, the horizontal component of the child's displacement will be positive.

The magnitude of the horizontal component of the child's displacement can be found by considering the right triangle formed by the horizontal line, the surface of the hill, and the vertical line. The angle between the horizontal and vertical lines is 90 degrees, and the surface of the hill, opposite 90-degree angle, is the hypotenuse. The angle between the horizontal and the hill is 27.2 degrees. The cosine of this angle, by definition, is the ratio of the lengths of the adjacent (horizontal) side and the hypotenuse:

`cos(27.2) = (adj)/(hyp)`

From here `Adj = (hyp)*cos(27.2)`

The length of the adjacent side is the magnitude of the horizontal component, and the length of the hypotenuse is the length of the hill: 20.1 m. So, the horizontal component is

20.1*cos(27.2) = **17.9 meters, rounded to the nearest tenth of a meter.**

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