Final velocity parallel to the track when the ski jumper lands

In summary, you used a -5,39 as the vertical velocity in post #11, but corrected it to -12,24 in post #12.
  • #1
KeevKK
11
0

Homework Statement


upload_2017-11-25_0-44-30.png

The skier leaves with 10 m/s
What is the final velocity parallel to the track when it is hit?

Homework Equations


The Attempt at a Solution


I have already found out d which is 21,27m and the time for the movement which is 1.6s.
I can also find the impact velocity, but how am I supposed to find the impact velocity parallel to the track?

I am sure that the solution is simple but I just have some conflicting thoughts regarding the method of finding it.

Thank you!
 

Attachments

  • upload_2017-11-25_0-44-30.png
    upload_2017-11-25_0-44-30.png
    13.1 KB · Views: 754
Physics news on Phys.org
  • #2
KeevKK said:
I am sure that the solution is simple but I just have some conflicting thoughts regarding the method of finding it.
And you need to share those thoughts with us so we can advise you about them.
 
  • Like
Likes KeevKK
  • #3
Okay sure.

These are my attempts:
upload_2017-11-25_11-21-15.png

In the first attempt i would know at what time the skier is 45 degrees and then I would put into the Vy equation, but it does not make sense to me since the movement is 1,6 seconds.

That is why I did the second attempt. But I am not sure if this is correct.

What do you think?
 

Attachments

  • upload_2017-11-25_11-21-15.png
    upload_2017-11-25_11-21-15.png
    111.2 KB · Views: 261
  • #4
You got about the right answer, but there is an easier way. Having found the horizontal and vertical components, find how much each of those contributes to velocity parallel to the plane and add them.
 
  • Like
Likes KeevKK
  • #5
Thank you! So you are saying that 20,15m/s is correct?

English is not my first language so I find it a little difficult to understand what you are saying.

When you say horizontal and vertical components you mean Vx and Vy right? And not Vxo and Vyo? I am a little confused of how to go about your easier solution.
 
  • #6
KeevKK said:
20,15m/s is correct?
I get slightly more.
KeevKK said:
Vx and Vy right?
Yes. What is the component of each of those parallel to the slope?
 
  • Like
Likes KeevKK
  • #7
What numbers did you use?

But I think I got it:
upload_2017-11-25_13-7-46.png

I still get 20,15m/s

Thanks a lot!
 

Attachments

  • upload_2017-11-25_13-7-46.png
    upload_2017-11-25_13-7-46.png
    86.4 KB · Views: 243
  • #8
I have a additional question for this problem. How do I derive a function for the curve when only Vinitial=10m/s is known?

Edit: Never mind, I solved it!:)
 
Last edited:
  • #9
Yet another question:

What if I want to find the center of curvature??

My thoughts are these: First use the formula Vf=Vi+a*t to find the tangential acceleration.
Then use this formula a^2=an^2+at^2 to find an which I will then use in the formula an=(v^2)/r in order to find r.

Is this the right method??
Thanks!
 
  • #10
KeevKK said:
I still get 20,15m/s
You seem to be quoting too many digits given the working along the way. Keep one more digit, at least, throughout the calculation than you intend to quote in the answer.
I did my own calc a bit more accurately and now get slightly under 20.
KeevKK said:
Vf=Vi+a*t to find the tangential acceleration.
I do not understand how that will give tangential acceleration.
You lnow the acceleration vector at any time. Just find the velocity direction and take the component of acceleration that is normal to it to get the radial acceleration.
 
  • Like
Likes KeevKK
  • #11
I intented to isolate a in the equation.

To find the center of curvature when t=0.9 I have done the following so far:
upload_2017-11-26_12-55-19.png

Then I found the velocity to be 15,431.

But is that my tangential acceleration as well?

I know the acceleration is 9.8m/s^2 but I am unsure what to do next.
 

Attachments

  • upload_2017-11-26_12-55-19.png
    upload_2017-11-26_12-55-19.png
    26.8 KB · Views: 680
  • #12
KeevKK said:
I intented to isolate a in the equation.

To find the center of curvature when t=0.9 I have done the following so far:
View attachment 215639
Then I found the velocity to be 15,431.

But is that my tangential acceleration as well?

I know the acceleration is 9.8m/s^2 but I am unsure what to do next.
You found the tangential velocity, not acceleration. Maybe that is what you meant.
Find the direction of the velocity and determine the component of the acceleration that is orthogonal to it. That is your radial acceleration.
 
  • Like
Likes KeevKK
  • #13
Thanks a lot for your help!

I have tried this:
upload_2017-11-27_13-51-39.png

What do you think?
 

Attachments

  • upload_2017-11-27_13-51-13.png
    upload_2017-11-27_13-51-13.png
    120.8 KB · Views: 417
  • upload_2017-11-27_13-51-39.png
    upload_2017-11-27_13-51-39.png
    82.1 KB · Views: 660
  • #14
KeevKK said:
Thanks a lot for your help!

I have tried this:
View attachment 215695
What do you think?
I don't understand... in post #11 you had a vertical velocity of -12,24, but here seem to have used -5,39.
 
  • Like
Likes KeevKK
  • #15
You are correct. I made a mistake. But was my approach correct?
 
  • #16
Got this instead after correction:
upload_2017-11-27_21-43-43.png
 

Attachments

  • upload_2017-11-27_21-43-43.png
    upload_2017-11-27_21-43-43.png
    76.2 KB · Views: 245
  • #17
upload_2017-11-28_20-31-34.png

This is the end result. Anyone care to confirm? :)
 

Attachments

  • upload_2017-11-28_20-31-34.png
    upload_2017-11-28_20-31-34.png
    37.7 KB · Views: 228

Related to Final velocity parallel to the track when the ski jumper lands

1. How is the final velocity of a ski jumper parallel to the track calculated?

The final velocity parallel to the track is calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken to reach the final velocity.

2. What factors affect the final velocity parallel to the track for a ski jumper?

The final velocity parallel to the track for a ski jumper can be affected by factors such as the initial velocity, the angle of the jump, air resistance, and the slope of the track.

3. How does air resistance impact the final velocity parallel to the track for a ski jumper?

Air resistance can significantly impact the final velocity of a ski jumper parallel to the track. As the ski jumper moves through the air, they experience air resistance, which can slow them down and decrease their final velocity.

4. Can the final velocity parallel to the track be greater than the initial velocity for a ski jumper?

Yes, the final velocity parallel to the track can be greater than the initial velocity for a ski jumper. This can happen if the ski jumper has a high initial velocity, a steep jump angle, and minimal air resistance.

5. How does the slope of the track affect the final velocity parallel to the track for a ski jumper?

The slope of the track can have a significant impact on the final velocity parallel to the track for a ski jumper. A steeper slope will result in a higher final velocity, while a flatter slope will result in a lower final velocity. The angle of the slope can also affect the landing of the ski jumper.

Similar threads

  • Introductory Physics Homework Help
2
Replies
40
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
226
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top