Finding distance along a ski-jump landing slope with horizontal takeoff velocity

[Lolagoeslala]

Nope. Accelerated motion. What's the equation of motion for an object undergoing acceleration? It should be in your notes or text.

There are two
one is

v2 - v1 / t = a

and the other

v2^2 - v1^2 / 2d = a

 

[gneill]

There are two
one is

v2 - v1 / t = a

and the other

v2^2 - v1^2 / 2d = a

Okay, those are fine if you happen to know the initial and final velocities, v1 and v2. But here you're interested in distance with respect to time; the only variables should be distance and time. y = ...

So, what equation of (accelerated) motion fits the bill?

 

[Lolagoeslala]

Okay, those are fine if you happen to know the initial and final velocities, v1 and v2. But here you're interested in distance with respect to time; the only variables should be distance and time. y = ...

So, what equation of (accelerated) motion fits the bill?

ok in that case it would be
2d/t^2 = a

 

[gneill]

ok in that case it would be
2d/t^2 = a

That'll do. Now make the distance variable y. Write it in the form y = ...

What symbol should you use in place of a for gravitational acceleration?

 

[Lolagoeslala]

That'll do. Now make the distance variable y. Write it in the form y = ...

What symbol should you use in place of a for gravitational acceleration?

y = gt^2/2

 

[gneill]

y = gt^2/2

Okay. Now, is y increasing or decreasing in value as time moves forward? (Is the skier rising or falling?)

 

[Lolagoeslala]

Okay. Now, is y increasing or decreasing in value as time moves forward? (Is the skier rising or falling?)

he is falling so there needs to be a negative sign infront of it

 

[gneill]

he is falling so there needs to be a negative sign infront of it

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]

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)

y(t) = - 1/2 g (x/v)^2

 

[gneill]

y(t) = - 1/2 g (x/v)^2

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?

 

[Lolagoeslala]

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?

y(x) = - 1/2 g (x/v)^2
x = -3/4

y(x) = - 1/2 g (-3/4v)^2
y(x) = - 1/2 g (-3/4v)^2
y(x) = 0.00689

 

[gneill]

y(x) = - 1/2 g (x/v)^2
x = -3/4 <---

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]

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]

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

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]

That looks better!

So that's the value of x where the curves intersect. Can you now find L, the distance along the slope?

ok so i know where they intersect which is 61.22
wouldn't the length be 61.22 too?

 

[gneill]

ok so i know where they intersect which is 61.22
wouldn't the length be 61.22 too?

Nope. That's the horizontal distance. You're looking for the distance along the slope. The 'L' in your diagram.

 

[Lolagoeslala]

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]

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

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]

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.

http://s1176.beta.photobucket.com/user/LolaGoesLala/media/Untitled_zps4a736718.jpg.html#/user/LolaGoesLala/media/Untitled_zps4a736718.jpg.html?&_suid=135745223970705186732013962768

like this?

 

[gneill]

http://s1176.beta.photobucket.com/user/LolaGoesLala/media/Untitled_zps4a736718.jpg.html#/user/LolaGoesLala/media/Untitled_zps4a736718.jpg.html?&_suid=135745223970705186732013962768

like this?

Yup. Be sure to indicate where the known angle is.

Alternatively, you could use geometry and similar triangles since you know that this triangle is similar to the 3-4-5 triangle from your initial diagram.

 

[Lolagoeslala]

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

 

[gneill]

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

That'll do it. But it's kind of taking the long way around

You can write $cos(\theta) = x/L$ so that $L = x/cos(\theta)$, and cos(θ) can be obtained from the similar 3-4-5 triangle as cos(θ) = 4/5.

Or, avoiding trig functions altogether, you can use the "raw geometry" of similar triangles:

$\frac{L}{x} = \frac{5}{4}$ so that $L = x \frac{5}{4}$

where x is 61.18 m, of course.

 

[Lolagoeslala]

THANK YOU SO MUCH FOR YOUR HELP!
u r like a god to me :D

 

[gneill]

THANK YOU SO MUCH FOR YOUR HELP!
u r like a god to me :D

Careful, my ego may explode

Glad I could help. Cheers!