If a skier jumps of the peak of a straight slope, when does it intersect?

[RoganSarine]

This is the only question in my physics section of projectiles I am having trouble with (This is ranked as the hardest question in this section of my text, please cut me some slack :P), because I have no idea how I would go about finding the intersection point of the skier with the slope. The reason I don't know how to do this is because I don't know how to turn the angle of depression of the slope versus the horizontal into something I can use.

http://img59.imageshack.us/img59/8373/picturen.jpg

A skilled skier knows to jump upwards before reaching a downward slope. Consider a jump in which the launch speed is vo = 10 m/s, the launch angle is theta = 9.0 degrees, the initial course is approximately flat, and the steeper track has a slope of 11.3 degrees. Figure 4-45a shows a prejump that allows the skier to land on the top portion of the steeper track. Figure -45b shows a jump at the edge of the steeper track. In Fig. 4-45a, the skier lands at approximately the launch level.

(A) In the landing, what is the angle phi between the skier's path and the slope? In Fig. 4-45b, (b) how far below the launch level does the skier land and (c) what is phi? (The greater the fall and greater phi can result in a loss of control in the landing)

Homework Statement

Skier:

Vi: 10m/s
Angle: 9.0 degrees
Vix: 10cos9.0
Viy: 10sin90

Track:

Slope: 11.3 degrees

Homework Equations

At this point, I really have no idea what I'm even doing.

The Attempt at a Solution

I can't even start because I don't know the time it would take for the skier to hit the ground, or the horizontal distance he would have.

Since the equation doesn't have a flat plain, I can't use the Range formula. Since I have no x and y coordinates, I can't really use the trajectory formula, and even if I could, how does it make up for knowing when I intersect and inclined plane? I don't know it's x and y components at all and have no way to break them up.

Also, (a) has an answer of 2.3 mainly because I assume that:

Since I think it's referring to Fig A, the skier would be landing at the same angle he launches at, and since the slopes angle is 11.3... The difference (phi) would be 2.3.

Anyone have a better explanation for (a)?

Anyone care to help me with (b) and (c)?

Further analysis:

What I wanted to try and do is use trajectory:

o = theta

y = (tan o)x - ((|g|*x^2)/(2vi^2*cos^2o))

I would try to be solving for the y value, but I don't know the x value at all, so I'd get stuck when trying to solve it anyway if I number crunch that down.

Again, I could solve for the x value EASILY if I knew time using the simple Vix*t = d, but I don't and can't find t out, so I can't solve for the horizontal distance. Also, since the plain isn't horizontal as I said before, i can't use the Range formula.

More work is below

 

[Jebus_Chris]

(a) is right. You go up, then back down to the same height so your y speed would be the same which would mean that your angle of velocity is the same.

For (b) the skier will land when the x and y displacement are equal to the slope.
[tex] \frac{y_f-y_i}{x_f-x_i} = \frac{\Delta y}{\Delta x}[/tex]
If you draw a little triangle on the incline you will find that she lands when...
[tex]tan\theta = y/x[/tex]

 

[RoganSarine]

Jebus, for (b) I can begin to see how the would work theoretically, and that is what I was trying to work at.

What I was thinking of doing (theoretically) is that I can assume that the slope of the ski hill is a linear equation that intersects the origin of where the skier begins.

Hence,

y = 11.3 degrees x + 0

And then use systems with the trajectory equation

y = (tan9)x - (9.8x^2)/(2(10cos9)^2)

=

y = .1584x - 9.8x^2/195

=

y = .1584 - .05x^2

Then use systems:

11.3 degrees * x = .1584 - .05x^2

The problem is... 11.3 degrees isn't a number I can use... Unless I put that into radians:

11.3(pi/180) = .197

therefore:

-.05x^2 - .197x + .1584

which gives me

x = 4.6, -.685 (Reject)

Substitute that back into the original trajectory equation to solve for y:

y = (tan9)(4.6) - (9.8(4.6)^2)/(2(10cos9)^2)

y = .729 - (207.368/195)

y = .729 - 1.06
y = -.33

Negative answer because of the negative displacement... This happens to be no where near the 1.4 m answer.
I've spent 2 hours at this question, and I would really like to do it before the morning, ha ha.

 

[Delphi51]

Hi RoganSerine,
I'm trying to do your problem just for fun (I'm a retired high school teacher).
I'm using the high school approach. The trajectory for (b) is determined by the vertical accelerated motion formula and the horizontal d = vt:
x = 10*cos(9)*t and y = -10*sin(9)t + .5*g*t^2
I've changed the sign on the y expression because I'm taking y to be the downward distance below the initial height.
On the slope, tan(11.3) = y/x.
If you put the x and y expressions for the trajectory into the expression for the slope you should get an expression with only the variable t so you can find the time the skier hits the slope. For this I get t = 0.721 seconds. But sometimes I make mistakes!
Putting this time back into the y expression gives y = 0.985 meters. This should be the answer for (b). I see it is not, but perhaps the method is okay and I have only made a mistake in calculating. Might be worth your while to check it out!

 

[Delphi51]

Yes, an error in the last calc for y. Now I'm getting 1.42 meters.

 

[RoganSarine]

Oh my gosh, thank you. I really try doing all these calculations using my high school formulas, but they don't really seem to work out right in some minor instances. Now with the concept of understanding what I have to do, I will go attempt it this way.

 

[RoganSarine]

...And then to attempt to find out what c is asking, which is I assume just finding the sin theta of the final velocity and compare the difference between the slope.

 

[Delphi51]

For (c) I think you must work with the skier's velocity components from the V = Vi + at for the vertical motion and the constant V for the horizontal motion.
The tan of the angle (with respect to horizontal) is equal to Vy/Vx evaluated at the time of landing.

 

[RoganSarine]

You are correct, I just got that. It took me a little while to realize I'd have to tan it, and I still don't QUITE get the algebra behind how that exactly works (I see it visually, but numerically I don't see it).

I'm also sick of looking at physics for taking 4 hours to do the 5 hardest questions of the section, so that may explain it.

 

[RoganSarine]

Wow, okay, yeah... I'm just really sick of physics. I get it -- wow... basic trig 101.

 

[Delphi51]

All this hard work is giving you EXPERIENCE - just what you need to do physics well!
It often helps to separate the big picture out from the details. In this problem you had the x and y functions of time for the trajectory and you had to figure out when that (x,y) position touched the slope, which had the linear equation y = x*tan(11.3) like good old y = mx+b. All three equations had to be true simultaneously, so you just solve them by substitution.

You'll be amazed at the problems you can do by the end of the year.

 

[RoganSarine]

Yeah... if you saw my giant work thing above, I had the right idea just I was using the trajectory equation to isolate x and y instead of the old tried-and-true formulas for x and y.

That's generally why I always the hardest 5 questions every section and normally skip the easy ones -- it definitely allows me to learn a few tricks to solve things.

Unfortunately, having a physics professor that can actually teach the course instead of just showing examples and crossing his fingers you can follow him is very hard to come by. I've only ever had one who could do it very well (and go figure that his course was the easiest one out of all my mega sciences and mathematics). Right now, with my current prof, I generally have to try and disregard or compartmentalize what he says to us because I don't agree with the way he works out his physics problems -- it's way more obscure than it needs to be.

 

[Delphi51]

Well, you can usually learn from any teacher but it is always a good idea to seek more than one viewpoint. Working a bit with classmates is great (as long as it doesn't go too far so you miss the struggles), PF is terrific and make sure you know the location of the grad students' common room.

I recall similar frustrations and one of the cures for me was reading Feynman's Lectures - I was inspired to look for clear and elegant ways of doing physics.

 

[RoganSarine]

Yeah, it's that this prof doesn't have the same way of doing physics that I do. I often tackle the problem in a completely different way which doesn't help me any.

Unfortunately, all the people who I used to bounce off my physics ideas and problems with are no longer in my class since they are finished with physics, so when I come to slightly obscure problems like this... It makes me have a very narrow scope on how to solve the problem.

 

[waba91]

I had to do the same problem and this helped. I guess we're using the same book this semester. Thanks for posting...

 

[RoganSarine]

Probably. The formula that Delphi made for us works 100% more accurately and better than the trajectory formula in Jearl Walker's Fundamentals of Physics. I've never got it to work right asides in the football question which is right after it (I believe that you can use this formula too for that question... I never tried it... But I did use this formula for every other questions on trajectory more or less -- it definitely made all the questions EXTREMELY easy).

Again, thanks Delphi :P

The rest of the book I havn't had trouble with since.

 

[zgozvrm]

(a) is right. You go up, then back down to the same height so your y speed would be the same which would mean that your angle of velocity is the same.

For (b) the skier will land when the x and y displacement are equal to the slope.
[tex] \frac{y_f-y_i}{x_f-x_i} = \frac{\Delta y}{\Delta x}[/tex]
If you draw a little triangle on the incline you will find that she lands when...
[tex]tan\theta = y/x[/tex]

The angle between the skier's path and the hill is 2.3 degrees when he lands just at the edge of the hill. This angle increases as his take-off point approaches the edge of the hill to approximately 17.862 degrees.

The shape of a trajectory is a parabola, such that its vertex is the high point (or peak) of the trajectory, and it lies directly above the focal point of the parabola (at an angle of 180 degrees above it). A line tangent to the peak of the parabola is oriented at 0 degrees. By definition, the angle of tangent lines along a parabola in this orientation, will become steeper and steeper the farther down the parabola you travel (approaching, but never quite reaching, an angle of 90 degrees ... the asymptote).

Therefore, the distance from the edge of the slope will determine how far down the slope the skier will make contact. The earlier the skier jumps, the sooner he will make contact with the slope (higher elevation), and the shallower his angle. The later the skier jumps, the farther down the slope he will come into contact with the slope, and the steeper his angle will be.

I also worked out the skier's landing angle with the slope when he takes off 2 meters before the hill: approximately 11.256 degrees.

 

[zgozvrm]

As it turns out, the angle the skier lands in relation to the slope is independent of his initial speed. It is dependent only on the angle of the slope, the angle of his take-off and the distance from the edge of the slope.