Angle of Elevation Stratton, VT Gondola
Over vacation, my family travelled to Stratton, VT to go skiing like we do every year. There are 11 lifts on this mountain, and a bunch of secret trails that lead to secret lodges if you can find them! This mountain is 3,875 feet tall, with the base of the mountain at height 1,872 feet with a vertical of 2,003 feet. The only lift that will take you from the base of the mountain to the summit is the gondola ride. My sister and I decided to take the ride up to determine the angle of elevation.
First things first; we gathered the information that we already knew: the height and the diameter of the mountain. The actual height that was being traveled was the height of the base subtracted from the height of the summit. First, I converted these measurements from feet to meters. 3,875 ft to 1,181 meters, and 2,003 ft to 571 meters. Subtract these two lengths and the resulting total height traveled equals 610 meters. The diameter of the mountain is 3599 ft, but from the base is 2896 ft (according to the Stratton Mountain website). The diameter at the base of the mountain is equal to 883 meters.
I timed the ride up the gondola and from the second it began moving to when we pulled into the lift exit, it took a total of 7 minutes and 57 seconds. This is equivalent to 477 seconds. I had to ask the man in the booth how fast the gondola traveled. He was very confused at this question and asked if I had motion sickness if I went too fast. He radioed someone in his booth and someone answered him that it was "around 5 mph if there are no stops." So with his strong confidence, I'm going to have to work with that. 5 miles per hour converts to 2.235 meters per second.
So here is the information that was gathered:
delta x: 883m delta y: 610m
VI: 2.235cos θ m/s 2.235sin θ m/s
VF: 2.235cosθ m/s 2.235sinθ m/sA: -9.8 m/s^2
T: 477 s
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Math
With all of this information, it was easy to use the formula: X=VT883 = 2.235cosθ (477)
883/477 = 2.235cosθ
833/477(2.235) = cosθ
cos^-1 = 35°
To double check this answer, I used the same formula but for Y instead: Y=VT
610 = 2.235sinθ (477)
610/477 = 2.235sinθ
610/466(2.235) = sinθ
sin^-1 = 34°
Conclusion:
In conclusion, the angle of elevation of the gondola is approximately 34.5°. The difference between the calculations in x and y can be attributed to error in measurement of either the precise diameter of the mountain from the base (which I am unsure how it was even calculated by the website of the mountain), or from lack of precision of the actual velocity the gondola was traveling. Overall, the two calculations are fairly close in answer and I can confidently assume that the angle of elevation is around 34.5°. If anyone only knew the mass of the gondola I would be able to calculate the tension on the rope, but I could not find that anywhere. I wonder at what angle of elevation the static friction will break and the gondolas won't be able to travel.... Stay tuned for the next blog (maybe we will find out!) Happy holidays and winter break to all.
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