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

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The discussion revolves around calculating 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. For the downward throw, the solution is straightforward using the formula, yielding a velocity of 26 m/s. However, the upward throw presents confusion regarding the sign of the initial velocity, as it must be treated as negative when considering the upward motion against gravity. Participants emphasize the importance of understanding the underlying physics principles, particularly how direction affects the signs in equations of motion. Clarifying these concepts is essential for solving similar problems in the future.
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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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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