Ski Jump Ramp: Calculating Gravitational Work, Total Speed, and Air Drag

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In summary, the skier leaves the ramp with a velocity of 24 m/s and returns to the ground at a point 14 m below the ramp with a total speed of 22 m/s. The gravitational force does work on the skier at the moment when he leaves the ramp, and the work done by the air drag is 888.5 Joules.
  • #1
King_Silver
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Homework Statement


A 72 kg skier leaves the end of a ski-jump ramp with a velocity of 24 m/s directed 25◦ above the horizontal. The skier returns to the ground at a point that is 14 m below the end of the ramp with a total speed of 22 m/s.

(a) What is the rate at which the gravitational force does work on the skier at the moment when he leaves
the ramp?

(b) With what total speed would he have landed if there were no air drag?

(c) What was the work done on the skier by the air drag?

Homework Equations



This is what I am uncertain about.

The Attempt at a Solution


[/B]
I think I was able to do part (c) by letting 1/2(m)(v2^2-v1^2) + mgh, Got a large number but it seems correct. What I am confused about is where to even start the first two questions, I have been looking through my notes but it seems I've misplaced my notes regarding this chapter which is fairly annoying. Could anyone recommend equations that would help resolve parts a and b?
 
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  • #2
Hello your majesty,

being uncertain about something is one thing. But you are uncertain about nothing: there is no relevant equation at all in the place where some are needed !

You know about potential energy from gravity ? something like mgh ? doing work often means changing energy. m and g don't change much, so it'll have to do with the change in h. What is the rate f change of h at the point of leaving the ramp ? :smile:
 
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  • #3
BvU said:
Hello your majesty,

being uncertain about something is one thing. But you are uncertain about nothing: there is no relevant equation at all in the place where some are needed !

You know about potential energy from gravity ? something like mgh ? doing work often means changing energy. m and g don't change much, so it'll have to do with the change in h. What is the rate f change of h at the point of leaving the ramp ? :smile:
m = 72kg
g = 9.81m/s^2
h = 14m

mgh = 9,888.5 Joules right? Would that be the rate of change when the skier leaves the ramp? :)
 
  • #4
No
1) the 14 m comes later.
2) a rate has the dimension of something per unit of time. Work is in Joules, rate of doing work is in Watt (Joules/second)

When leaving the ramp, what is the rate of change of h ?
 
  • #5
BvU said:
No
1) the 14 m comes later.
2) a rate has the dimension of something per unit of time. Work is in Joules, rate of doing work is in Watt (Joules/second)

When leaving the ramp, what is the rate of change of h ?

Rate of change is the slope or the first differential with respect to h right? That is my understanding of ROC anyway. Is it the rate of change in terms of time?
like dh/dt?
Or instead of dh/dt... dQ/dt? (Q = theta = 25 degrees)
 
  • #6
It is in terms of time. It is dh/dt. :smile:

But: dQ/dt doesn't help (25 degrees is 25 degrees, right?). Make a little drawing for that scary moment. See how high the skier is at that moment and how fast his/her height increases with time .
 
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  • #7
BvU said:
It is in terms of time. It is dh/dt. :smile:

But: dQ/dt doesn't help (25 degrees is 25 degrees, right?). Make a little drawing for that scary moment. See how high the skier is at that moment and how fast his/her height increases with time .

I feel so stupid, this is what a week of a common cold can set you back with. I understand that but I can't seem to see the relation to the question! I feel like I'm missing some sort of information here, or something key. It is probably a very easy thing too if its something to do with rate of change.
 
  • #8
Flattery will get you nowhere. If you throw something in the air with a vertical speed of 10 m/s, its height initially changes at a rate of ... 10 m/s !
 

Related to Ski Jump Ramp: Calculating Gravitational Work, Total Speed, and Air Drag

What is a ski jump ramp?

A ski jump ramp is a structure used in the sport of ski jumping. It consists of a takeoff ramp, a table, and a landing slope. The ramp is designed to provide the skier with enough speed and lift to jump off the table and cover a long distance in the air.

How is a ski jump ramp built?

A ski jump ramp is typically built on a hillside or mountain slope. It is constructed using wooden planks and steel beams, and the shape and angle of the ramp are carefully calculated to ensure safe and efficient jumps. The ramp is also covered with a layer of snow to create a smooth and fast surface for the skiers.

What are the dimensions of a ski jump ramp?

The dimensions of a ski jump ramp vary depending on the level of competition. However, the International Ski Federation (FIS) has set certain standard dimensions for ski jump ramps used in Olympic and World Cup events. These dimensions include a minimum takeoff height of 2.5 meters and a maximum landing slope angle of 37 degrees.

How do ski jumpers adjust their technique on different ramps?

Ski jumpers must adapt their technique based on the size and angle of the ramp. On larger ramps, they need to generate more speed and lift to cover a greater distance. On smaller ramps, they must focus on maintaining a more aerodynamic position to achieve maximum distance. They also need to adjust their timing and balance to account for any variations in the ramp's shape or structure.

What safety precautions are in place for ski jump ramps?

Ski jump ramps are designed and maintained with safety as a top priority. The ramps are regularly inspected and repaired to ensure they are in good condition. There are also safety nets and padding placed at the bottom of the landing slope to cushion any potential falls. Ski jumpers also wear specialized equipment, such as helmets and suits, to protect themselves during jumps.

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