016 10.0 points
One of the great dangers to mountain climbers is an avalanche, in which a large mass of snow and ice breaks loose and goes on an essentially frictionless “ride” down a mountainside on a cushion of compressed air.
The acceleration of gravity is 9.8 m/s2 .
30.5◦
Ifyouwereona30.5◦slopeandan avalanche started 303 m up the slope, how much time would you have to get out of the way?
I tried to use the summation of the forces equals mass times acceleration formula but that got me 4.973875957 as the acceleration and then I plugged that in to d=Vit+1/2at^2 but that got me the wrong answer I don't know what else to do
To solve this problem, we need to consider the forces acting on the avalanche and use the principles of Newton's second law.
First, let's analyze the forces involved. The only force acting on the avalanche is gravity, which can be decomposed into two components: one parallel to the slope and the other perpendicular to the slope. The component parallel to the slope will cause the avalanche to accelerate down the slope, while the perpendicular component will not affect its motion along the slope.
The parallel component of gravity can be found using trigonometry. The angle of the slope is given as 30.5°. So, the component of gravity acting parallel to the slope is given by:
Fpar = mg * sin(30.5°)
Where:
Fpar = force parallel to the slope
m = mass of the avalanche (which we don't have, but we don't need it to solve for time)
g = acceleration due to gravity (9.8 m/s^2)
Now, using Newton's second law (F = ma), we can equate this force to the product of mass and acceleration:
mg * sin(30.5°) = m * a
Simplifying, we have:
a = g * sin(30.5°)
a = 9.8 m/s^2 * sin(30.5°)
This value of acceleration (a) will be used in the equation of motion to find the time it takes for the avalanche to reach a distance of 303 m down the slope.
To find the time, we can use the equation of motion:
d = Vit + (1/2)at²
where:
d = distance traveled (303 m)
Vi = initial velocity (0 m/s, since the avalanche starts from rest)
a = acceleration (g * sin(30.5°))
t = time
Plugging in the known values, we get:
303 m = 0*t + (1/2)(9.8 m/s^2 * sin(30.5°)) * t²
Simplifying further, we have a quadratic equation:
4.9 * sin(30.5°) * t² = 303
Now, we can solve this equation for t.
t² = 303 / (4.9 * sin(30.5°))
t ≈ sqrt(61.84)
t ≈ 7.86 seconds (rounded to two decimal places)
Therefore, you would have approximately 7.86 seconds to get out of the way of the avalanche.