a skier is acceleration down a 30.0 degree hill at 3.50 m/s^2. how long (in seconds) will it take her to reach the bottom of the hill. assuming she starts from rest and accelerates uniformly, if the elevation change is 120m?

the distance down slope is 120/sin30 = 240m

So, just solve
1.75t^2 = 240

To solve this problem, we need to use the equations of motion for uniformly accelerated motion.

We are given:
Acceleration (a) = -3.50 m/s^2 (negative because the skier is accelerating down the hill)
Angle of the hill (θ) = 30.0 degrees
Elevation change (Δy) = 120 m
Initial velocity (u) = 0 m/s (since the skier starts from rest)

First, we need to find the acceleration in the direction parallel to the incline. Since the acceleration is acting down the hill, we need to find the component of the acceleration parallel to the incline.

Acceleration parallel to the incline (a_parallel) = a * sin(θ)

a_parallel = -3.50 m/s^2 * sin(30.0 degrees)
a_parallel ≈ -1.75 m/s^2

Now we can use the equation of motion to find the time it takes for the skier to reach the bottom of the hill.

Δy = u * t + (1/2) * a_parallel * t^2

Plugging in the known values, we have:

120 m = 0 * t + (1/2) * (-1.75 m/s^2) * t^2

Rearranging the equation, we get:

0.875 * t^2 = 120

Divide both sides by 0.875:

t^2 = 137.143

Taking the square root of both sides, we get:

t ≈ 11.71 seconds

Therefore, it will take her approximately 11.71 seconds to reach the bottom of the hill.