# Projectile motion of a snowmobile

1. Sep 23, 2011

### getty102

1. The problem statement, all variables and given/known data

The snowmobile is travelling at 10m/s when it leaves the embankment at A. The angle that the snowmobile leaves its elevated jump is 40 degrees. Determine the time and flight from A to B and the range R of the trajectory.
The slope of the mountain is a 3-4-5 triangle.

2. Relevant equations
y=y_0+10sin40(t)-.5(9.81)t^2
R=10cos40(t)

3. The attempt at a solution
I wasn't sure how to get the height, range, and time(in the air) using only the equations of projectile motion.

2. Sep 23, 2011

### omoplata

I am guessing from what I read that B is either up a mountain from A or down a mountain, and the slope of the mountain is in the form of a 3:4:5 right angled triangle.

Is B higher than A, or lower than A?

Is y/R equal to 3/4, or is it equal to 4/3 ?

Without knowing these the question can't be answered. Is there a diagram that comes with this problem?

Last edited: Sep 23, 2011
3. Sep 23, 2011

### getty102

You are correct, the point A starts at an elevated position while point B is at a lower position. The 3-4-5 triangle has its 3 side underneath point A and the 4 side would be considered the Range.

4. Sep 23, 2011

### omoplata

OK, then if you consider the motion upwards to be positive, since the vertical position of the projectile ends up lower than it's starting vertical position (i.e. since it falls), your first equation should change to this.

-y=y_0+10sin40(t)-.5(9.81)t^2

Your second equation is correct for the range,

R=10cos40(t)

Also,

y_0 = 0

From the sides of the 3:4:5 triangle we can get the following relation,

y/R = 3/4

These four equations are all you need to find y, R and t.