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##u## is not a function of ##t##. ##u## is the (constant) initial value at ##t =0##.Differentiate it said:Well even if i specify v there's u = f(t), which I can still graph, it's a function!
##u## is not a function of ##t##. ##u## is the (constant) initial value at ##t =0##.Differentiate it said:Well even if i specify v there's u = f(t), which I can still graph, it's a function!
How? Also pls answer the question in post, #66, it'll be really helpfulPeroK said:##u## is not a function of ##t##. ##u## is the (constant) initial value at ##t =0##.
I think you meant ##u=f(T)## not ##f(t)##. Remember,there's a difference between ##t## and ##T##.Differentiate it said:Well even if i specify v there's u = f(t), which I can still graph, it's a function!
By definition. I'm using the notation:Differentiate it said:How?
I haven't been keeping track or t and T, sorry. But at this point, just answer my question in post #66, and if the answer is no, please explain why. I'll be very thankfulPeroK said:By definition. I'm using the notation:
$$v(t) = u + at$$where ##v(t)## is the time dependent velocity and ##u = v(0)## is the initial velocity.
Note that although ##v## is a continuous function of ##t##, we also tend to use ##v## for the "final" velocity. There is, therefore, a notational asymmtery between ##u##, which is assumed to be fixed, and ##v##, which is assumed to be a variable end-point.
That's the minimum speed needed to touch the balloon because with any speed less than that it wouldn't touch the balloon.Delta2 said:Ok I see now, basically the *intuitive* claim here is that the minimum speed is such that when the bullet reaches the balloon it has the same speed as the balloon. How do we justify this claim more formally?
Hm ok but the approach described at post #2 solves the problem without the need to state this claim.Mister T said:That's the minimum speed needed to touch the balloon because with any speed less than that it wouldn't touch the balloon.
It's assumed.Delta2 said:Hm ok but the approach described at post #2 solves the problem without the need to state this claim.
No it isn't assumed. You just say there $$h=25t+125$$ (height of balloon), $$h'=v_0t-0.5gt^2$$ (height of bullet) form the quadratic equation (with respect to t) $$h=h'$$ and the minimum requirement for ##v_0## pops in a "wonderous " way from the requirement that the discriminant $$(v_0-25)^2-4\cdot 0.5\cdot g \cdot 125\geq 0$$of the quadratic is greater or equal to zero.Mister T said:It's assumed.
That's the key. Without understanding that, you are just asking fairly irrelevant questions about minimising functions in general.Differentiate it said:I haven't been keeping track or t and T, sorry.
I'm sorry, but please just answer my question from post #66. That's it.PeroK said:That's the key. Without understanding that, you are just asking fairly irrelevant questions about minimising functions in general.
The substitution ##v = u - gt## didn't change the nature of the equations, but replaced the variable ##v## by ##u## and ##t##, because we wanted to solve for the impact time in terms of ##u##. Having both ##v## and ##u## in the equation was not what we wanted.
The solution to that equation is ##t = T##, which leads to a function for ##u## and ##T##, which we can minimise. Note that ##u## is a function of ##T## but not of ##t##.
I don't understand your question. It makes no sense to me. I've explained why I used ##v = u - gt##. We are not minimising SUVAT equations. How many times do you have to be told that?Differentiate it said:I'm sorry, but please just answer my question from post #66. That's it.
Some functions, like linear ones, like u = 10t + 25, don't have a minima.(i got this from v = u +at where i fixed the value of v to 25 m/s) Other functions like 125/t + 25 + 5t do have a minima. So, instead of fixing v to a certain value, if we substitute v = sqrt(u^2 + 2as) for v in v = u +at where s = 25t + 125, and rearrange for initial velocity u, we get u = 125/t + 25 + 5t, which does have a minima. So, my final question was, do we substitute in something different for v(in this case it was sqrt(u^2 + 2as)) instead of a fixed value like 25 to ensure that the function 𝘩𝘢𝘴 𝘢 𝘮𝘪𝘯𝘪𝘮𝘢. I don't think this has anything to do with t or T, which I didn't keep track of, I'm sorry, but please, just answer this questionPeroK said:I don't understand your question. It makes no sense to me. I've explained why I used ##v = u - gt##. We are not minimising SUVAT equations. How many times do you have to be told that?
It looks like that's what you did, so yes. But I would use the word "create" rather than "ensure".Differentiate it said:So, my final question was, do we substitute in something different for v(in this case it was sqrt(u^2 + 2as)) instead of a fixed value like 25 to ensure that the function 𝘩𝘢𝘴 𝘢 𝘮𝘪𝘯𝘪𝘮𝘢.
Thanks! And, sorry for my poor wordingMister T said:It looks like that's what you did, so yes. But I would use the word "create" rather than "ensure".
I still don't understand how i would find such a function (if I knew why some functions didn't have a minima(due to physical reasons), i would know what type, scratch that, which equation to use)Differentiate it said:Thanks! And, sorry for my poor wording