Basketball thrown straight up from the ground

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The discussion revolves around understanding the kinematic equations related to the motion of a basketball thrown straight up. Participants seek clarity on how to apply these equations to the specific scenario presented, despite the equations being blurred. There is a request for a clear explanation of the relevant kinematic equations and their adaptation to this situation. Additionally, there is a call for responses to be formatted in LaTeX for better readability. The aim is to accurately describe the physics of the ball's vertical motion.
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Homework Statement
Basketball thrown straight up from the ground, Vo = 5m/s, what equations correctly finds position y(t), velocity of the basketball vy(t), and acceleration
Relevant Equations
kinematic equations
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Would greatly appreciate an explanation on this question, not sure I even understand what they're asking.
 
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I assume that the scribblings in red are your attempt at a solution.

They are giving you three equations, which are too blurred to read, and are asking you to find which one(s) correctly describe(s) the physical situation of throwing the ball straight up in the air.

You say that the relevant equations are "the kinematic equations." What are they and how would you adapt them to this particular situation? Please post the answers in a legible form, ideally using LaTeX. Click on the link "LaTeX Guide" (lower left) to learn how.
 
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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