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Height question

  1. Jan 14, 2008 #1
    Hey guys, im new here, i was wondering if you could help me solve the following problems
    ok...A stone is thrown vertically upward at a speed of 48.70 m/s at time t=0. A second stone is thrown upward with the same speed 4.390 seconds later. At what time are the two stones at the same height?
    now i know the height where they meet is: 7.159s, but i cant solve the other part:
    At what height do the two stones pass each other?
    What is the downward speed of the first stone as they pass each other?

    any help would be appreciated
  2. jcsd
  3. Jan 14, 2008 #2


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    Use s = ut + 1/2at^2 and solve the first.
    As for the second, you calculate a maximum point for y then work from then on.

    Just try and solve for a function of the displacement of each one and then differentiate to get velocity results.
  4. Jan 14, 2008 #3
    ut?? or do you mean at?
  5. Jan 14, 2008 #4


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    Science Advisor
    Homework Helper

    In these equations 'u' is initial velocity.
    'v' is final velocity, 's' is displacement, 't' is time and 'a' is acceleration.

    Another equation that is useful for the second part is:
    v^2 = u^2 + 2 a s

    Be careful to get the signs of u,v and a correct.
  6. Jan 14, 2008 #5
    ok but how do i find 'u'??
  7. Jan 14, 2008 #6
    You've already said what it is ...
  8. Jan 14, 2008 #7
    but once it reaches the peak... it fall down with accelration = 9.8m/s^2 right?
  9. Jan 14, 2008 #8
    Be careful with your signs. If "+" is up, then the grav. accel. is negative. Otherwise, yes, the acceleration due to gravity applies consistently throughout.
  10. Jan 14, 2008 #9
    wow im so lost, lol... this is pathetic... is there anyway i can compute a parabola?
  11. Jan 14, 2008 #10
    Look at the equation for height (s) in Post #2: it gives s as a 2nd order polynomial in t, i.e. a parabola. How's that? Think about that and maybe try plotting it (height as a function of time). Once you can do that for the two stones, you should be well on your way.
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