One-Dimensional Kinematics: Height vs Time

  • Thread starter rdn98
  • Start date
  • #1
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This problem has a picture. You can view it at http://www.geocities.com/rockdog_84/One.htm

Ok. Maybe I'm missing something here, but for part one, I thought free-fall acceleration would be 9.8 m/s^2. Of course, computer didn't take it, so is there a way to find it from the graph?

As for the second part of the question, I thought initial velocity is 0, but its not the right answer. :-(
I guess another way to find initial velocity would be take the derivative of x(t) when t = 0, but I don't know the equation for this problem.

Any insight is appreciated.
 

Answers and Replies

  • #2
Claude Bile
Science Advisor
1,471
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9.8 m/s^2 is the correct acceleration due to gravity on Earth, however the scenario in question takes place on the planet Zoot. If a question does not take place on Earth, then that is a big red flag that acceleration due to gravity is NOT 9.8 m/s^2.

You can figure out g by using s = 1/2 at^2 + u*t (Take the second half of the graph where u is 0).

Once you know g, you can use u^2 = v^2 - 2*a*s to find the final velocity. (Take the first half of the graph this time so v = 0)
Note: -
u = initial velocity
v = final velocity
t = time
a = acceleration
s = displacement

Claude.
 
  • #3
39
0
big thanks!

Man, I would never have thought of that! Thank you so much.
 

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