- #36
gneill
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Lolagoeslala said:y = gt^2/2
Okay. Now, is y increasing or decreasing in value as time moves forward? (Is the skier rising or falling?)
Lolagoeslala said:y = gt^2/2
gneill said:Okay. Now, is y increasing or decreasing in value as time moves forward? (Is the skier rising or falling?)
Lolagoeslala said:he is falling so there needs to be a negative sign infront of it
gneill said:Good.
Now, you've got your two parametric equations for the trajectory:
##x(t) = v_x t##
##y(t) = -\frac{1}{2}g t^2##
Can you combine them to write y in terms of x? (that is, eliminate t in the equation for y)
Lolagoeslala said:y(t) = - 1/2 g (x/v)^2
gneill said:Alright. But no y(t) any more. It's y(x). The t has been eliminated. That's the equation of the parabola that the skier follows.
So now you have the equation of the line representing the ski slope and the parabola representing the skier's path. Can you find their intersection?
No, that's not right; x is not a constant. Where's your equation for the line? Look back to posts #23 and #24.Lolagoeslala said:y(x) = - 1/2 g (x/v)^2
x = -3/4 <---
Lolagoeslala said:y =- 3/4x
y(x) = - 1/2 g (x/v)^2
-3/4x = -1/2g(x/v)^2
3/4x = 1/2g(x/v)^2
3/4x = 1/2gx^2/v^2
2 x 3/4 x v^2 / g= x
600/g = x
61.22 = x
gneill said:That looks better!
So that's the value of x where the curves intersect. Can you now find L, the distance along the slope?
Lolagoeslala said:ok so i know where they intersect which is 61.22
wouldn't the length be 61.22 too?
Lolagoeslala said:Ok i see so itll be like this..
horizontal distance = 61.22 m
L = ?
Angle = 37°
cos 37° = a/61.22 m
cos 37°x 61.22m = a
and the L is 48.89 m
gneill said:Bit of a problem with your working there. Consider,... is the hypotenuse of a triangle ever shorter than either of the other sides? Here L is the hypotenuse of a triangle with one side of length 61.22 m. Draw your triangle with the 'x' side and 'L' hypotenuse in place and redo the calculation.
Lolagoeslala said:http://s1176.beta.photobucket.com/user/LolaGoesLala/media/Untitled_zps4a736718.jpg.html#/user/LolaGoesLala/media/Untitled_zps4a736718.jpg.html?&_suid=135745223970705186732013962768
like this?
Lolagoeslala said:Okay so what i was thinking was i could find the side that is opposite to the angle
so like
tan 37 = o / 61.22 m
o = 46.13 m
and use a^2 = b^2 + c^2
a^2 = (61.22m)^2 + (46.13m)^2
a= 76.6 or 77
Lolagoeslala said:THANK YOU SO MUCH FOR YOUR HELP!
u r like a god to me :D