What Is the Maximum Height Reached by a Falling Brick?

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The discussion centers on calculating the maximum height reached by a brick that falls from a crane at a velocity of 5 m/s from a height of 6 meters. The initial assumption is that the greatest height is simply 6 meters, but there is uncertainty regarding the brick's velocity at the moment it separates from the load. Participants explore the implications of the brick's initial velocity and the effects of gravity on its trajectory. Clarification is sought on whether the brick's downward velocity contributes to a greater height before it begins to fall. The conversation emphasizes the need to consider both the initial height and velocity to determine the maximum height accurately.
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A load of bricks is being lifted by a crane at a velocity of 5 m/s when one brick falls off 6m above the ground. What is the greatest height the brick reaches above the ground?

How do you solve it?

The attempt at a solution
Since it fell of 6 meters above the ground, the greatest height it reached was 6 meters.

But I'm not sure. Am I missing something?
 
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What velocity did the brick have just before it separated?
 
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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