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Ski jump-energy

  • #1

Homework Statement


A skier (m=53.00 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 8.00 m and D = 9.50 m, find H.
http://img517.imageshack.us/img517/616/prob21a.gif [Broken]


Homework Equations


.5mv^2+mgh=.5mv^2+mgh


The Attempt at a Solution


I first divided to vertical distance by 9.81 to get .81seconds, then 9.50/.81 to get the horizontal velocity, which is 11.72 m/s

I then plugged the information into the above equation for

53*9.81*x=.5*53*11.72^2+53*9.81*8
solving for x i got 15.01 m, which is not correct.
 
Last edited by a moderator:

Answers and Replies

  • #2
38
0
Well, I think you need to consider the velocity of the skier when he leaves the ramp. mgH-mgh is the kinetic energy at that point, right? And if there is no drag, then the horizontal velocity at the time he leaves the ramp will be constant throughout the drop.

Then you can use kinematics to solve for that x velocity, right? Use a system of kinematics equations(I think this is what you are missing) to get rid of time and I believe there is only one variable left: H

EDIT: I get 10.82m for the answer. If that isn't correct, then the way I suggested might not be right.
 
Last edited:
  • #3
Well, I think you need to consider the velocity of the skier when he leaves the ramp. mgH-mgh is the kinetic energy at that point, right? And if there is no drag, then the horizontal velocity at the time he leaves the ramp will be constant throughout the drop.

Then you can use kinematics to solve for that x velocity, right? Use a system of kinematics equations(I think this is what you are missing) to get rid of time and I believe there is only one variable left: H

EDIT: I get 10.82m for the answer. If that isn't correct, then the way I suggested might not be right.
your answer was correct, thanks for the help but could you please tell me how you used kinematics to solve for velocity? after plugging the hight u gave me back into the energy equation i see the velocity was 7.44, which is way off from what i was getting.
 
  • #4
38
0
Yeah, of course.
So in the x direction, I used the equation:
[tex]
\Delta x = v_{x}t+\frac {1}{2}at^2
[/tex]
So in terms of this problem:
[tex]
D= v_{x}t
[/tex]
Because there is no acceleration.

In the y direction, I used the same equation, and in terms of this problem:
[tex]
h= \frac {1}{2}at^2
[/tex]
I left off the vt because there is no initial y velocity since the skier leaves the ramp horizontally.

I solved this for t and substituted it into the first equation, and then solved it for vx

I think you can finish it once you have this velocity.
 

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