Motion in two or three dimensions

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The discussion revolves around calculating the maximum launch speed of a projectile to avoid entering a temperature inversion layer at height h, using the equations of motion and conservation of energy. Initially, the user attempted to solve the problem using kinematic equations but struggled to reach the correct answer, which is given as (2gh)^(1/2). After guidance, they realized that applying conservation of energy simplifies the solution significantly. The conservation of energy approach shows that potential energy at height h equals kinetic energy at launch, leading to the same result with fewer steps. The user appreciates the insight into the energy relationship, which clarified the problem.
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Homework Statement


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.


Homework Equations


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


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.
 
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Think about it from a conservation of energy perspective, how does the initial kinetic energy relate to the maximum height?
 
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.
 
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.
 
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.
 
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