the distance down slope is 120/sin30 = 240m
So, just solve
1.75t^2 = 240
So, just solve
1.75t^2 = 240
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.