Question

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.

Answer 1

Answer:

20 m/s

Explanation:

First, we calculate for the acceleration down the incline using $a = gsin\theta$.

a = (9.80)sin(20°)

a = 3.35

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

$v_{i}$ = 0 m/s

a = 3.35

$\Delta s$ = 62 m

$v_{f} = \sqrt{2(3.35)(62)}$

$v_{f} =20.381 = 20 m/s$