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Homework Help: Motion in two or three dimensions

  1. Aug 11, 2010 #1
    1. The problem statement, all variables and given/known data
    A projectile is being launched from ground level with no air resistance. You want to avoid having it enter a temperature inversion layer in the atmosphere a height h above the ground.(a) What is the maximum launch speed you could give this projectile if you shot it straight up?Express your answer in terms of h and g.

    2. Relevant equations
    According to me the equation that applies is h(t)=Yi+Vit-(1/2)gt2

    3. The attempt at a solution
    I'm lost in this problem I tried doing it as an optimization problem from Calc. I but it just didn't make sense. The other thing I did was solve for Vi in the above equation but it din't give me anything close to the answer in the book I got Vi=(h+4.905t2)/t and the answer in the back is (2gh)(1/2). Any help will be appreciated, Thanks.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 11, 2010 #2
    Think about it from a conservation of energy perspective, how does the initial kinetic energy relate to the maximum height?
  4. Aug 11, 2010 #3
    Ok I think I got it, can someone please let me know if this is right or not. This is what I did.

    1.(Eq.a)h=Vt-(1/2)gt^2 and from a previous equation found (Eq.b)t=(2h/g)^(1/2)

    2.From Eq.a V=[h+(1/2)gt^2]/t

    3.Plugging (b) into new (a) gives V=2h/(2h/g)^(1/2)

    4.Rationalizing gives V=[2h(2h/g)^(1/2)]/(2h/g)

    5.And after solving the division V=g*[(2h/g)^(1/2)]

    6.And by properties of exponents V=(2hg)^(1/2) which is the answer in the back of the book.

    Any comments will be appreciated in case I did something wrong, Thanks.
  5. Aug 11, 2010 #4
    Yeah, no, that looks right. But seriously you should really think about this in terms of conservation of energy. It will take 2 lines, max.
  6. Aug 11, 2010 #5
    Yeah I did it the other way and is way simpler using conservation of energy, oh well now I have two ways of doing it, thanks for the help.

    1. a)PE=mgh b)KE=(1/2)mv^2

    2. By conservation of energy PE=KE so mgh=(1/2)mv^2

    3. m divides out on both sides and you get gh=(1/2)v^2

    4. Therefore v=(2gh)^(1/2) which is what i got the other way but way shorter.

    I would have never thought about energy being related in this problem on my own thanks zhermes.
    Last edited: Aug 11, 2010
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