Stone Throw, Found answer but don't understand how ?

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SUMMARY

The discussion centers on the physics problem of determining the velocity of a stone thrown from a bridge with an initial vertical velocity of 4.0 m/s after 2.2 seconds, considering both upward and downward directions. The participant successfully calculated the downward velocity as 2.6 x 101 m/s using the formula Vf = Vi + Aav(Δt). However, confusion arises regarding the use of the same equation for both scenarios, particularly the sign change for initial velocity (Vi) when the direction is reversed. Understanding the underlying principles of acceleration due to gravity and vector direction is essential for grasping this concept.

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Homework Statement


A stone is thrown from a bridge with an initial vertical velocity of magnitude 4.0m/s/ Determine the stone's velocity after 2.2s if the direction of the initial velocity is (a) upward, and (b) downward. Neglect air resistance.

Homework Equations


Aav=Vf - Vi / Δt

Vi + Aav(Δt) = Vf

The Attempt at a Solution



b) 2.6 x 101m/s
Easy to find as you enter in the given information into the formula.

(a) on the other hand I have difficulty understanding. I attempted the problem, with this formula
(-Vi) + Aav(Δt) = Vf. The answer I got corresponds to the answer in text, but I don't really understand how that works. I'd like to know for similar problems in the future.
 
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You use the same equation to solve both problems, but if v_i is positive in b, it must be negative in problem a.
 
willem2 said:
You use the same equation to solve both problems, but if v_i is positive in b, it must be negative in problem a.

Yes that's apparently obvious, I want to understand why however. The "theory" behind it.
 

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