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A ski hill is angled 20° above the horizontal. If a skier starts from rest at the top of the hill, how fast will she be traveling after she has traveled 62 m 62 ⁢ m down the slope? Assume that friction is negligible.

Sagot :

Answer:

20 m/s

Explanation:

First, we calculate for the acceleration down the incline using [tex]a = gsin\theta[/tex].

a = (9.80)sin(20°)

a = 3.35

Then, we use one of the kinematics equations, [tex]{v_{f}}^{2} = {v_{i}}^{2} + 2 a\Delta s[/tex].

[tex]v_{i}[/tex] = 0 m/s

a = 3.35

[tex]\Delta s[/tex] = 62 m

[tex]v_{f} = \sqrt{2(3.35)(62)}[/tex]

[tex]v_{f} =20.381 = 20 m/s[/tex]