1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating gravity from height and time of a jump

  1. Jul 5, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm trying to calculate a value for gravity based on a known jump height and time. It's working when the take-off and landing heights are the same but I'm struggling for a solution if the landing point is higher than the take-off. I'm more interested in understanding a solution method than getting an answer to a specific question, so I've used symbols instead of values, thanks.

    An astronaut lands on a planet and jumps up to a height of 'S' and lands 'T' seconds later on a rock at height 'R'. What is the value of the planet's gravity?

    2. Relevant equations

    s = ut + 0.5*at2
    v = u + at

    3. The attempt at a solution

    When take-off and landing are the same, I use s = ut + 0.5*at2 for the falling time (half total time), with u=0, rearranging s = 0.5*at2 to give an answer a = 2s / (t/2)2 I think this method is fine.

    With a height difference, I guess a solution is based on the same equation, but separating the jump into two parts - the initial upward part to reach the rock height on the way up, and then treating the the rest of the jump as before. But I don't seem able to figure out how much time is spent in each part of the jump?
     
  2. jcsd
  3. Jul 5, 2011 #2

    Delphi51

    User Avatar
    Homework Helper

    Welcome to PF, Fizz!
    Interesting problem! I wrote Vi for the initial upward velocity and t for the time to maximum height so I have 3 unknowns including g. That means 3 equations needed. I'm an old high school teacher so I usually only remember two accelerated motion formulas: d = Vi*t + .5*a*t and V = Vi + a*t.
    I used the d one twice for distance R and distance T. And the V one for the maximum height when V = 0. It should be just a matter of eliminating t and Vi to get an answer for g. Be careful about the signs; I took acceleration to be -g so g comes out positive, but you could do it with g negative if you prefer.
     
  4. Jul 9, 2011 #3
    Thanks for your suggestion, sounds much simpler than the quadratic equation I ended up with working out how much time was spent going up and how much coming down!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook