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:
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.
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...
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!:)
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...
Okay sure.
These are my attempts:
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...
Homework Statement
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...