Solve Freefall Problem: Find Speed of Stone Thrown Upward

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SUMMARY

The problem involves calculating the initial speed of a stone thrown straight upward to reach a height of 20.0 meters. The final speed is determined to be 19.8 m/s using kinematic equations. The relevant equations include v = u + at, s = ut + 1/2at², and v² = u² + 2as. The symmetry of projectile motion simplifies the calculation, allowing the use of the equation v² = 2gh to find the initial speed directly.

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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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