# Speed of a Snowmobile: Snow Jump

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## 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 = 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 obtained by
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*_{oy} = (34.5744)^{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

No condition is permanent.