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From what height will two objects hit the ground at the same speed?

  1. Jul 24, 2012 #1
    1. The problem statement, all variables and given/known data

    There are 2 objects both with mass 0.50 kg which are dropped from rest at height h. One is dropped on Earth and experiences air resistance and one is dropped on moon and experiences no air resistance. The gravitational acceleration on earth is 9.8 and on moon it is 1.6.

    From what height h will the objects hit the ground at the same speed?

    2. Relevant equations
    F=-kv for air resistance
    k=0.17

    3. The attempt at a solution

    I know that we will somehow use the gravitational acceleration.
     
  2. jcsd
  3. Jul 24, 2012 #2
    Well I'd start by figuring out the equations that will tell you how long it will take for each object to fall (it'll be a function of h). Then I'd set the two equal to each other.

    Do you see where you would get those equations?
     
  4. Jul 24, 2012 #3
    Unfortunately, no I don't see where to get the equations because if I set it equal the h will cancel out.
     
  5. Jul 24, 2012 #4
    Taking down as positive.
    [itex]\ddot {y}=-b\dot y+g[/itex]
    or
    dv/dt=-bv+g
    This is a differential equation of motion.
     
    Last edited: Jul 24, 2012
  6. Jul 24, 2012 #5
    Hello,

    I am looking for at what height h will the velocity of the two objects be constant. I don't see how this while help solve. Also, I don't know what each of the variables stands for.
     
  7. Jul 24, 2012 #6
    You can use SUVAT equation for the height of the motion on the moon.
    With air resistance of the earth, its a bit complicated to find the height with differential equation involved.
     
  8. Jul 24, 2012 #7
    I am sure that there is another way to do this because we have not learned about SUVAT or differential equations.
     
  9. Jul 25, 2012 #8
    The time it takes for an object to fall from a height H is [itex]\sqrt{\frac{2h}{g}}[/itex]

    It should be easy to set two sides of an equation equal to each other (moon and earth) and solve for h.
     
  10. Jul 25, 2012 #9
    Im not sure I understand because when I make
    [itex]\sqrt{\frac{2h}{g}}[/itex] = [itex]\sqrt{\frac{2h}{g}}[/itex]

    then won't h cancel out? also, you haven't taken into account the air resistance on earth?
     
  11. Jul 25, 2012 #10
    Oh I totally misread the original question, I'm sorry for misleading you with my stuff. I thought wanted something different.

    I can think of two ways I'm pretty sure I could solve this with (Energy stuff, and calculus stuff), but neither factors in air resistance, so I'll need to think about it for a tiny bit.

    Again, sorry for talking about the wrong stuff.
     
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