Where Would a Wooden Ball Accelerate When Thrown Up and Then Let Fall Down?

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When a wooden ball is thrown upwards and then allowed to fall, it experiences constant acceleration due to gravity throughout its motion. The acceleration remains non-zero as long as the ball is in the air, calculated as a = Fgrav/m. Only when the ball is held or when the ground exerts a normal force does the net force become zero, resulting in no acceleration. Thus, the ball accelerates downwards during its entire free fall. Understanding this principle is crucial for solving problems related to motion under gravity.
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Homework Statement


If you were to throw a wooden ball up one meter and then let it fall down to a negative meter, where would it accelerate?


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All the way down.
Gravity always acts on it, so as long as it is in the air it has non-zero acceleration a = Fgrav/m.
Only when you hold it (straining your muscles to do so) or letting the ground exert a normal force, you can cancel it such that the net force is zero hence a = 0..
 
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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