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[BlackSalad]

now I assure you I only turn to GAF when I have no other options. I have tried everything to my knowledge to solve this problem, perhaps you can help me again (sorry for this)

A skier leaves the ramp of a ski jump with a velocity of v = 8.0 m/s, = 11.0° above the horizontal, as in Figure P4.67. The slope is inclined at 50.0°, and air resistance is negligible. (Assume up and right are positive, and down and left are negative.)

a) Find the distance from the edge of the ramp (take-off point) to where the jumper lands.

its asking for the hypotonuse, not the delta x (was sort of wordy)
I figured that the Vi it gives me is not the Velocity of the x components, so I used 8 * sin(11) to find the x component velocity.

theta 2 = 11 degrees + 50 degrees

now what I've done:


(for xf):
1)Solved for t in the equation:
Vix * t = d cos(theta2)

2)substituted the t from step 1 into the equation:
yi = -1/2(g)(t^2) = -d sin(theta2)

3) solved for d (the hypotonuse of that right triangle)


I got 46.8332meters which is not the correct answer. I have also tried it using the 8.00 m/s as the Velocity x component, and got 48.6027 meters which was also incorrect.

 

[Dilbert]

There are a couple of ways to do this problem. I don't have a lot of time right now, but I'll give you two ideas to start with:

1) The "classic" way of doing this problem is to transform the parametric equations of motion into a Cartesian equation. Remember that, in the absence of air resistance, the flight path of an object with some velocity component in the y-direction will be a parabola. If you can plot that parabola on a coordinate system and find the point where it intersects the line formed by the mountain slope, you've solved the problem. In case you don't remember parametric -> Cartesian, remember that you can write the 2-D equations of motion as x = v0xt and y = v0yt - gt². Using the first one, you can solve for t to get t = x/(v0x) and plug in that value in the second equation to get something in the form y = f(x) which should be quadratic in x. The equation for a line, obviously, is y = mx + b.

2) The other way of doing the problem -- which is somewhat more complicated -- involves rotating the coordinate system. If you rotate the picture by 50 degrees so that the downward slope is now along the x-axis, then the problem LOOKS more like the traditional sort of problem. The steps for solving it will be familiar -- calculate the time of flight from the y-axis kinematic equation, plug in that time into the x-axis kinematic equation and voila. However, there is a complication -- when you decompose the gravitational acceleration vector (g), there will be components in both x and y. This means that the velocity along the x-axis will NOT be constant.

It looks like you were trying to do approach #2, but forgot to decompose g along the new coordinate axes.

Hope this helps...

 

[rareside]

Are you the resident physics guru, jinx? Man, I majored in physics in college, but now I'm jaded as hell and don't want to think about physics for a few years.

I'm glad someone's here to help though...

I hated Cartesian equations... Give me some chaos theory, that's the good stuff!