Solve Freefall Problem: Find Speed of Stone Thrown Upward

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To solve the problem of a stone thrown upward to a height of 20.0m, the relevant kinematic equations are needed. The final speed can be calculated using the equation v^2 = u^2 + 2as, where 's' is the displacement and 'a' is the acceleration due to gravity. The symmetry of the motion indicates that the speed at the peak height can be found using v^2 = 2gh, where 'h' is the height. The initial speed required to reach this height is approximately 19.8 m/s. Understanding these equations is essential for solving similar problems effectively.
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I need help getting the base equations for a situation like this so I can complete all of the other problems on the page. Thanks ahead of time.

Homework Statement


A stone is thrown straight upward and it rises to a height of 20.0m. With what speed was it thrown?


Homework Equations


This is what I need.


The Attempt at a Solution


Answer: 19.8 m/s (given, the equations are what I need)
 
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g is constant so the kinematic equations apply

those are v=u+at;s=ut+1/2at^2;v^2=u^2+2as

Where v=final speed,u=initial speed,a=acceleration,s=displacement
 
It's always very useful and time saving to use the symmetry in the rise and fall of an object. This means that in this case we have to calculate the speed after a fall of 20 m from rest. The eqn v^2 = 2gh gives it directly.
 
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