Sandbag Dropped from Moving Balloon

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A hot air balloon ascends at 10 m/s when a passenger drops a sandbag from 40 m above ground. The acceleration due to gravity is -9.81 m/s², and the initial velocity of the sandbag matches the balloon's upward speed. The equation used to determine the time in the air is v1 = x0 + v0Δt + 0.5aΔt². After solving, it is confirmed that the sandbag stays in the air for approximately 2.86 seconds before hitting the ground. The discussion emphasizes the importance of using the correct initial velocity and omitting negative time values.
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Homework Statement


Hey guys, I'm struggling on the following question:

Consider a hot air balloon traveling upwards at a constant speed of 10 m/s. A passenger drops a small sandbag over the side of the balloon when the balloon was 40 m above the ground. For how long does the sandbag stay in the air before falling to the ground?
Known:
a=-9.81 m/s2
x0=40 m
x1=0 m
v0=0 m/s
v1=?
t=?

Homework Equations


v1=x0+v0Δt+0.5aΔt2

The Attempt at a Solution


0=40+0.5(-9.8)Δt2
-40=-4.9Δt2
8.16=Δt2
Δt2=±2.86

Since time cannot be negative, it takes 2.86 seconds, is this correct? Thanks in advance.
 
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Lycouras said:
Consider a hot air balloon traveling upwards at a constant speed of 10 m/s.
...
v0=0 m/s

Since the balloon is moving upwards and the sandbag was dropped from the balloon, the initial velocity of the sandbag is the same as the velocity of the balloon. Other than that, you did everything right.
 
So from there you use the quadratic formula to solve for t, then again omit the negative answer? Thank you!
 
correct
 
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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