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!

2x distance

  1. Feb 19, 2008 #1
    1. The problem statement, all variables and given/known data
    two students are standing on a fire escape, one twice as high as the other. simultaneously, each drops a ball. if the first ball strikes the ground at time t1 when willt he second ball strike the ground?
    A)t2=4t1
    B)t2=5/4t1
    C)t2=2t1
    D)t2=squareroot of 2 times t1


    2. Relevant equations



    3. The attempt at a solution
    i tried plugging in numbers and well its two times the height so wouldn't it be C??
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 19, 2008 #2
    Yay I get to use my skydiving experience! When jumping out of an airplane I know it'll take roughly 10 seconds to fall the first 1000 feet, then roughly 5 seconds for every subsequent 1000 feet. So 10 seconds for 1000 feet, 15 seconds for 2000 feet(pretty gross estimation, but it overestimates how far you've fallen usually which is better than underestimating for obvious reasons!)

    So think about the equation you should be using here

    d=-1/2*g*t^2

    What you've said in thinking it's C is that if I have a ball that falls a distance D1 in T1 seconds then if T2=2*T1, then D2=2*D1

    is that true?
     
  4. Feb 19, 2008 #3
    wen i tried pluging in numbers i kept the initial velocity, aceleration the same and made one distance x and the other one 2x and then i tried finding the final velocity. then using one of the formulas i found how much time it would take and compared them. so the Vi and A is the same but the D and Vf are different.
     
  5. Feb 19, 2008 #4
    You'll confuse yourself trying to plug in numbers

    so D=-1/2*g*T^2, 1/2 and g are constants

    Now we're saying it falls twice D in time T2, so 2*D=-1/2*g*T2^2, and you have said that T2=2*T, or that's what choice C says anyways. So what you're saying is 2*D=-1/2*g*(2*T)^2

    You KNOW that D=-1/2*g*T^2 from our very first step, so if you solve the bolded equation for D, do you get the same thing?
     
  6. Feb 19, 2008 #5
    yea the both equation do equal out. but how would that prove that the time would be 2 times if the distance is 2x?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: 2x distance
  1. Distance ? (Replies: 4)

  2. 10 -2x = 90 - x (Replies: 1)

  3. How to graph 2x+y=1 (Replies: 3)

Loading...