Slope and Projectile Problem: Finding Distance Covered on a Landing Slope

[Lolagoeslala]

Homework Statement

Eddie "The Eagle" Edwards takes off on the 90 m ski-jump with a horizontal velocity of 20 m/s. If the gradient of the landing slope is as shown, find the distance L along the slope covered by his jump,ignore any effect of the air on this eagle.

Image given:
http://s1176.beta.photobucket.com/user/LolaGoesLala/media/yeahok_zps7743bc20.png.html

The Attempt at a Solution

So, they have given us the horizontal velocity which is 20 m/s [ E ]
and i am not sure what the 90 m represent.

Is it the vertical length or is the diameter along the base of the triangle.

 

[jedishrfu]

He started from zero velocity travelledsome dist then jumped with a horizontal vel of 20m/sec and the diagram shows the angle of the skijump.

Since he fell some distance and gained kinetic energy you should have enough info to solve the problem.

 

[gneill]

I think that you can ignore the "90 m"; consider it a name of the jump.

Concentrate on the fact that Eddie is launched horizontally, and that you are given information about the gradient (the "slope" of the slope, so to speak).

 

[haruspex]

From a little Googling, the 'size' of a ski jump is to do with distance from take-off to some standard landing line, rather like par for a golf hole. If so, it doesn't supply any useful info here. So it would seem the question ought to say "with a velocity of 20 m/s horizontal", i.e., there is no vertical velocity initially. (In practice, the jump slope is a little below horizontal.)

 

[Lolagoeslala]

Okay.
so i know the horizontal velocity = 20 m/s
θ = E 37 S

so i think i would be able to use the conservation of energy in this question.
mgh = 1/2mv^2
gh = 1/2v^2
(9.8m/s^2)h = 1/2(20m/s)^2
h = 20.4 m

so can i use sin 37 = 20.4 m / L

 

[gneill]

The 20 m/s is a horizontal velocity. The "h" in "mgh" pertains to a vertical displacement. So they are unrelated. Conservation of energy isn't going to cut it here.

Instead, look at Eddie's trajectory as a function of horizontal displacement and the line of the ski slope as a function of horizontal displacement.

 

[Lolagoeslala]

The 20 m/s is a horizontal velocity. The "h" in "mgh" pertains to a vertical displacement. So they are unrelated. Conservation of energy isn't going to cut it here.

Instead, look at Eddie's trajectory as a function of horizontal displacement and the line of the ski slope as a function of horizontal displacement.

well i was looking over my notes.
and you are right.
But i also found this equation
distance horizontal = v^2 sin 2θ / g
but doesn't that only work for like the velocity that includes both vertical and horizontal

 

[gneill]

well i was looking over my notes.
and you are right.
But i also found this equation
distance horizontal = v^2 sin 2θ / g
but doesn't that only work for like the velocity that includes both vertical and horizontal

That would be the range equation for a projectile launched with speed v at angle θ over a horizontal surface. That, too, doesn't pertain to the given situation. I doubt that you'll be able to pull a "canned" formula out of your pocket to fit this situation. Instead, you'll have to develop a method of solution.

If you step back from the problem and look at your diagram, you should be able to see that there are two functions and a point of their intersection that you're interested in. One is the parabolic trajectory of the skier, the other is the straight line of the ski slope.

If you set your origin at the launch point, can you write an equation expressing the line of the hill as a function of horizontal distance? How about the vertical position of the skier as a function of horizontal distance?

 

[Lolagoeslala]

I am sorry but i am confused by your two question.

 

[gneill]

Two functions that have a point of intersection.

 

[Lolagoeslala]

So your first question is about the slope (blue line) and the x distance ?

 

[gneill]

So your first question is about the slope (blue line) and the x distance ?

Yes. Write an equation for it given the information that you have.

 

[Lolagoeslala]

So like...
well the distance is just east while the blue is east 37 south to the distance horizontally

 

[gneill]

So like...
well the distance is just east while the blue is east 37 south to the distance horizontally

Hmm. Have you taken a math course in functions? Equations of lines, parabolas, and so on?

 

[Lolagoeslala]

Hmm. Have you taken a math course in functions? Equations of lines, parabolas, and so on?

yeah. but isn't this physics thought? :O

 

[gneill]

yeah. but isn't this physics thought? :O

The language of physics is mathematics. It provides the tools you need to solve problems.

Here you have an opportunity to use what you learned in your functions course to solve a physics problem. All the math you've learned will come in handy as you go forward. The trick is to recognize its applicability to a given problem

 

[Lolagoeslala]

The language of physics is mathematics. It provides the tools you need to solve problems.

Here you have an opportunity to use what you learned in your functions course to solve a physics problem. All the math you've learned will come in handy as you go forward. The trick is to recognize its applicability to a given problem

Oh i see..
But i don;t get it I am i supposed to use the y =mb+x functions?

 

[gneill]

Oh i see..
But i don;t get it I am i supposed to use the y =mb+x functions?

Yup. The ski slope can be represented by a straight line. The trajectory of the skier is a parabola. They intersect at a point. Physics is essentially the world described in terms of mathematics.

 

[Lolagoeslala]

okay so if they have a point of intersection that would mean they must be equal to one another so.

if the two equations are:
y = mx

and the other one is y = -ax

combing them would be

mx = -ax

right?

 

[gneill]

okay so if they have a point of intersection that would mean they must be equal to one another so.

That is correct

if the two equations are:
y = mx

and the other one is y = -ax

combing them would be

mx = -ax

right?

Not quite. One is a straight line (y = mx), but the other one is a parabola.

For the first you have to find the slope m. For the second, you need to find the equation of the parabola that matches the trajectory of Eddie.

Hint: The equations of motion for a projectile are those of a parabola in what is called "parametric form", where time "t" is the parameter. You will have to convert this to Cartesian form (y as a function of x).

 

[Lolagoeslala]

That is correct

Not quite. One is a straight line (y = mx), but the other one is a parabola.

For the first you have to find the slope m. For the second, you need to find the equation of the parabola that matches the trajectory of Eddie.

Hint: The equations of motion for a projectile are those of a parabola in what is called "parametric form", where time "t" is the parameter. You will have to convert this to Cartesian form (y as a function of x).

umm does this go with the Grade 12 curriculam, the second part? and how would i find the slope if i don't have the exact information about the graph :(
isn' t there another way to solve this?

 

[gneill]

umm does this go with the Grade 12 curriculam, the second part? and how would i find the slope if i don't have the exact information about the graph :(
isn' t there another way to solve this?

While I'm not familiar with your particular curriculum, I would assume that if you've taken functions then you've seen them in parametric form: x = f(t); y = g(t), so that x and y are individual functions of the parameter t. The equations of motion of the projectile are such a pair of functions, where the y-component and x-component of the trajectory are both functions of time (t).

The information to find the slope of the ski hill is certainly on the initial drawing that you posted (look at the the little triangle formed at the bottom right).

I can't think of another way that's any simpler; it uses tools that I think you must have seen and used at the grade 12 level (although as I said, not having seen your curriculum I might be making an error of judgement there. But I can't think of another approach that relies on any simpler mathematical tools).

 

[Lolagoeslala]

okay so for the straight line it is..

Slope of hypotenuse = length of vertical arm / length of horizontal arm.
Slope of hypotenuse = 3/4 is the slope

so its y =3/4x for the straight line

now i need help with the parabola shape.

 

[gneill]

okay so for the straight line it is..

Slope of hypotenuse = length of vertical arm / length of horizontal arm.
Slope of hypotenuse = 3/4 is the slope

so its y =3/4x for the straight line

Make that -3/4 for the slope, since it's sloping downwards; For every 4 positive units of x forward, the curve drops by 3.

now i need help with the parabola shape.

Start by writing the usual equations of motion for the x and y components of the trajectory.

 

[Lolagoeslala]

Make that -3/4 for the slope, since it's sloping downwards; For every 4 positive units of x forward, the curve drops by 3.

Start by writing the usual equations of motion for the x and y components of the trajectory.

like what do you mean by break it into x and y component or writing their equation like would u like to expand on that, it would really help me.

 

[gneill]

When you have a projectile thrown in a constant gravitational field, you consider the vertical motion and horizontal motion separately. The horizontal motion has a constant speed (so x = v_x*t). The vertical motion is accelerated motion, the acceleration being given by the gravitational constant, g. So write the two equations using the information that you have on hand.

Hint: You were told that Eddie takes off horizontally with a given speed --- no initial y-component of his velocity.

 

[Lolagoeslala]

When you have a projectile thrown in a constant gravitational field, you consider the vertical motion and horizontal motion separately. The horizontal motion has a constant speed (so x = v_x*t). The vertical motion is accelerated motion, the acceleration being given by the gravitational constant, g. So write the two equations using the information that you have on hand.

Hint: You were told that Eddie takes off horizontally with a given speed --- no initial y-component of his velocity.

x = v_x*t

and for the Y = 0

right?

 

[gneill]

x = v_x*t

Yes.

and for the Y = 0

Nope. y should be a function of time, not a constant value. Sure, it starts at zero at the launch point, but after that it goes negative, accelerating downwards (just like a dropped object, right? After all, it is a dropped object!).

 

[Lolagoeslala]

Yes.Nope. y should be a function of time, not a constant value. Sure, it starts at zero at the launch point, but after that it goes negative, accelerating downwards (just like a dropped object, right? After all, it is a dropped object!).

Ok so for Y = - (v2 - v1 /t)

 

[gneill]

Ok so for Y = - (v2 - v1 /t)

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