Attempted Part I & II Problems

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The discussion centers on solving parts i and ii of a physics problem related to projectile motion. The user is struggling with part ii, specifically how to reach the x value found in part i without an air resistance term. It is clarified that without air resistance, the problem simplifies to basic projectile motion. Additionally, a key point is made regarding the evaluation of an integral, emphasizing that the constant c2 should be set to zero before integration to avoid invalid assumptions. The conversation highlights the importance of correctly applying mathematical principles in physics problems.
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Homework Statement
A bullet of mass m is shot vertically up with initial speed v0. It is subject to air resistance that is quadratic in speed (constant of proportionality c2).

i. What is the maximum height reached?

ii. What initial speed u0, would be necessary to reach the same maximum height in the absence of air resistance?

Express your answers in terms of c2, g, m, and v0.
Relevant Equations
For part i: m(dv/dt) = -c2(v^2) - mg
I have an attempted solution to part i here. I'm primarily having trouble with part ii since I don't see how the x found in part i can be attained if there's no air resistance term.
 

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theredbarronmvr said:
I'm primarily having trouble with part ii since I don't see how the x found in part i can be attained if there's no air resistance term.
If there is no air resistance, then it's a simple projectile motion problem.
 
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PeroK said:
If there is no air resistance, then it's a simple projectile motion problem.
Ahh of course, thank you! I was definitely overthinking it.
 
theredbarronmvr said:
I have an attempted solution to part i here. I'm primarily having trouble with part ii since I don't see how the x found in part i can be attained if there's no air resistance term.
In case you're wondering, when you evaluated the integral, you assumed ##c_2 \ne 0##, so it isn't valid to set ##c_2=0## in the expression you derived. (I assume that's the difficulty you were running into.) You have to set ##c_2## to 0 before you integrate.
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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