Speed of a Snowmobile: Snow Jump

The Physics Factbook
Edited by Glenn Elert -- Written by his students
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The Problem

animated gifThis is a mechanics problem that a physics student should be able to solve

View the clip of the snowmobile flip at killsometime.com.

In the video, a snowmobile is jumped off of an incline and flips backwards, then lands upright on the ground. Assuming that the snowmobile took off at a 45° angle and that it landed on ground at the same height as that from which it took off, you should be able to determine:

  1. Peak Altitude
  2. Range
  3. Takeoff Speed

1. Altitude

First an appropriate formula must be chosen. The formula y = vot + ½at2 is the appropriate one in this case.

In order to determine the total time that the snowmobile is in the air, we can count the frames in the video from take-off to landing, then divide by the frame rate of the video, which was 30 frames/second.

The formula chosen above contains vo. In order to know the value of vo, we must use only the time from the peak to the ground where vo is 0, which in this case is exactly half the total time due to the 45 degree take-off angle. This means that t is equal to ½ of the total time.

total time = frames / frame rate
total time = 36 frames / 30 frames/second
total time = 1.2 seconds
t = 1.2 seconds / 2 = .6 seconds

y = vot + ½at2
y = (0 m/s)(.6 s) + ½(9.8 m/s2)(0.6 s)2 = 1.764 m

2. Range

To find range, s, the horizontal velocity of the snowmobile must first be found.

To find horizontal velocity in this example, initial vertical velocity must first be found, and then a horizontal component can be found using trigonometry.

vfy2 = voy2 + 2asy
0 = voy2 + 2(-9.8 m/s2)(1.764 m)
voy = (34.5744)½ = 5.88 m/s

vector diagram
x = (5.88 m/s) / (tan 45°) = 5.88 m/s

s = vxt
s = (5.88 m/s)(1.2 s) = 7.056 m

3. Take-off Speed

vector diagram
x = (5.88 m/s) / (cos 45°) = 8.32 m/s

Matthew Grabczynski -- 2005

Physics on Film pages in The Physics Factbook™ for 2005
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