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Well, the trajectory of the skiier will be parabolic, based on acceleration due to gravity and his speed off the end of the ramp. To solve this, I would find the equation describing the parabola, and then find the intersection of that parabola and the linear equation describing the slope. Then, use that intersection and the linear equation to find the distance.
Problem is that I haven't used those math skills for a long long time. I guess once you have both equations, you could set them equal to each other, subtract one equation from both sides, and solve for zero to get the intersection. (There will be two zeros, but only one will be relevant, the other will be before he leaves the ramp).
Also, I might be mistaken, but I don't think the potential energy at the bottom is mg*4.4. The ramp levels off horizontally, so you might assume that all the potential energy at the top has been converted to kinetic enegy.
I was wrong in my last post, you probably calculated the velocity correctly.
Now I'm getting 116.4 with the new number, but anyway:
you've got your Y position equation, Y=Yo + Vot + 1/2*at^2. In this problem, you can set Y=0 because that's the point you want to land at, you set your initial position Yo as 4.4 + S*sin(30), and you plug in 9.8 for a. vo is equal to zero because you have no initial velocity on the y axis. Thus, your equation should look like:
0 = 4.4 + S*sin(30) - 4.9t^2
For your x position equation (which is the same format), your initial X position is 0 and you have no horizontal acceleration. Your final X position will be S*cos(30). Therefore, your equation should look like this:
S*cos(30) = (28.2)*t
Now you've got two equations with two unknown variables, S and t. Solve the x equation for t, plug that into your y equation, and then use the quadratic formula to solve for S.
Also with these hill problems, if you want you can think like this, you can run the axis parallel to the slope, sorta like you do with block's sliding
You know the slope's angle with the horizontal, so what happens is your initial velocity is no longer perfectly horizontal, and gravity doesn't act perpendicular to the axis, all that does is make a perfectly normal problem with an initial velocity that's not horizontal and the weird addition of gravity not going straight "down" in your reference frame
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Problem is that I haven't used those math skills for a long long time. I guess once you have both equations, you could set them equal to each other, subtract one equation from both sides, and solve for zero to get the intersection. (There will be two zeros, but only one will be relevant, the other will be before he leaves the ramp).
I don't want to post my solution if I messed up that badly, but 120m sounds a little odd too. That's a really long distance and death worthy fall.
edit: oh, duh, you gave us the numbers
edit again: oh wait no you didn't, what's the mass of the skier
Now I'm getting 116.4 with the new number, but anyway:
you've got your Y position equation, Y=Yo + Vot + 1/2*at^2. In this problem, you can set Y=0 because that's the point you want to land at, you set your initial position Yo as 4.4 + S*sin(30), and you plug in 9.8 for a. vo is equal to zero because you have no initial velocity on the y axis. Thus, your equation should look like:
0 = 4.4 + S*sin(30) - 4.9t^2
For your x position equation (which is the same format), your initial X position is 0 and you have no horizontal acceleration. Your final X position will be S*cos(30). Therefore, your equation should look like this:
S*cos(30) = (28.2)*t
Now you've got two equations with two unknown variables, S and t. Solve the x equation for t, plug that into your y equation, and then use the quadratic formula to solve for S.
My quad looks like this:
0 = - 4.9(t^2) + 12.12t + 4.4
Your quad is
0 = - 4.9(t^2) + 2*(12.12)t + 4.4
You know the slope's angle with the horizontal, so what happens is your initial velocity is no longer perfectly horizontal, and gravity doesn't act perpendicular to the axis, all that does is make a perfectly normal problem with an initial velocity that's not horizontal and the weird addition of gravity not going straight "down" in your reference frame