Sandbag Dropped from Moving Balloon

[Lycouras]

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.

 

[frogjg2003]

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.

 

[Lycouras]

So from there you use the quadratic formula to solve for t, then again omit the negative answer? Thank you!

 

[frogjg2003]

correct

 

[Nickie]

Yes, your solution is correct. The sandbag will take approximately 2.86 seconds to fall to the ground. This can be found by using the equation v1=x0+v0Δt+0.5aΔt2 and solving for Δt. It is important to note that this is an approximation since air resistance and other factors may affect the actual time it takes for the sandbag to fall.