# Speed of a Snowmobile: Snow Jump

## The Problem

This 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:

- Peak Altitude
- Range
- Takeoff Speed

## 1. Altitude

First an appropriate formula must be chosen. The formula y = v_{o}t + ½at^{2} 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 v_{o}. In order to know the value of v_{o}, we must use only the time from the peak to the ground where v_{o} 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* = *v _{o}t* + ½

*at*

^{2}

*y*= (0 m/s)(.6 s) + ½(9.8 m/s

^{2})(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.

*v _{fy}*

^{2}=

*v*

_{oy}^{2}+ 2

*as*

_{y}0 =

*v*

_{oy}^{2}+ 2(-9.8 m/s

^{2})(1.764 m)

*v*= (34.5744)

_{oy}^{½}= 5.88 m/s

*x* = (5.88 m/s)/(tan 45°) = 5.88 m/s

*s* = *v _{x}t*

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

## 3. Take-off Speed

- To determine take-off speed, components and trigonometry can once again be used.

*x* = (5.88 m/s)/(cos 45°) = 8.32 m/s

Matthew Grabczynski -- 2005

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