if you are on a 29.2 degree slope and an avalanche started 236 m up the slope how much time would you have to get out of the way
A bit longer than it takes an object to fall 236 m, straight down. That would be sqrt(2H/g) = 4.9 seconds. An exact solution would be rather complicated. A pressure wave will go through the snow and an air pressure wave will eventually form in front of the advancing front. Pressure waves may speed up the avalanche, while the falling snow is ed by surface snow ahead of it.
The teacher may expect you to assume something like frictionless motion down the slope, but it isn't that simple.
You may find this previous answer interesting:
http://www.science-facts.com/weather-climate/avalanche-speeds-explained/
To calculate the time you would have to get out of the way, we can use the equation of motion for an object sliding down an inclined plane:
\(s = ut + \frac{1}{2}at^2\),
where:
- \(s\) is the vertical distance traveled by the object (236 m in this case),
- \(u\) is the initial velocity of the object (0 m/s as it starts from rest),
- \(a\) is the acceleration of the object due to gravity (\(9.8 \, \text{m/s}^2\) for Earth),
- \(t\) is the time taken.
Since the object moves vertically down the slope, we need to consider the component of gravity along the slope. The effective acceleration due to gravity acting along the slope can be calculated as:
\(a_{\text{eff}} = a \sin(\theta)\),
where \(\theta\) is the angle of the slope (29.2 degrees in this case).
Now, by substituting the given values into the equation and solving for \(t\), we can find the time you would have to get out of the way:
\(236 = 0 \cdot t + \frac{1}{2} \cdot a_{\text{eff}} \cdot t^2\).
Let's calculate it step-by-step:
Step 1: Convert the slope angle from degrees to radians:
\( \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \).
\( \theta_{\text{rad}} = 29.2 \times \frac{\pi}{180} \).
Step 2: Calculate the effective acceleration due to gravity:
\( a_{\text{eff}} = a \cdot \sin(\theta_{\text{rad}}) \).
\( a_{\text{eff}} = 9.8 \times \sin(29.2 \times \frac{\pi}{180}) \).
Step 3: Rearrange the equation to solve for \( t \):
\( t = \sqrt{\frac{2s}{a_{\text{eff}}}} \).
\( t = \sqrt{\frac{2 \times 236}{9.8 \times \sin(29.2 \times \frac{\pi}{180})}} \).
Now, let me calculate it for you.
To determine the time you have to get out of the way, we can use the equations of motion. Let's break down the problem step by step:
1. Determine the vertical distance (height) covered by the avalanche:
The height (h) covered by the avalanche can be calculated using trigonometry. Since the slope is given as 29.2 degrees, we can use the sine function:
h = 236 m * sin(29.2°)
2. Calculate the time it takes for the avalanche to reach the bottom:
We'll use the equation of motion for vertical displacement:
s = ut + (1/2)gt^2
where s is the vertical displacement (h), u is the initial velocity (0 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time we need to find. Note that we ignored the deceleration caused by air resistance.
Rearranging the equation, we have:
h = (1/2)gt^2
2h = gt^2
t^2 = (2h)/g
t = √((2h)/g)
3. Substitute the value of h into the equation to calculate the time (t):
t = √((2 * 236 m * sin(29.2°)) / 9.8 m/s^2)
4. Calculate the final answer:
Evaluating the above expression, we can find the time it takes for the avalanche to reach the bottom of the slope, which is the time you have to get out of the way.
Please note that this calculation assumes a constant downward acceleration due to gravity and neglects factors like air resistance, variation in avalanche speed, and other real-world complexities. It's always crucial to prioritize your safety and seek expert advice or follow local guidelines when dealing with avalanche situations.