How long does it take for a package to fall from a hot-air balloon?

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Homework Help Overview

The problem involves a package dropped from a hot-air balloon that is ascending at a rate of 13 m/s from a height of 90 m. Participants are exploring the time it takes for the package to reach the ground and the speed at which it impacts the ground.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial velocity of the package, questioning whether it should be considered as 0 or 13 m/s. They explore different kinematic equations to solve for the time and final velocity of the package.

Discussion Status

Some participants have confirmed their calculations for part (a) of the problem, while others are still discussing the implications of their results for part (b). There is an ongoing exploration of whether the final velocity should be expressed as negative due to the direction of motion.

Contextual Notes

Participants are operating under the assumption that the acceleration due to gravity is acting on the package after it is dropped, and there is a consideration of sign conventions in their calculations.

suxatphysix
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Homework Statement


A hot-air balloon is ascending at the rate of 13 m/s and is 90 m above the ground when a package is dropped over the side.

(a) How long does the package take to reach the ground?


(b) With what speed does it hit the ground?



Homework Equations






The Attempt at a Solution



Tried 2a[tex]\Delta[/tex]y=Vf[tex]^{2}[/tex]-vi[tex]^{2}[/tex]

I don't know if the initial velocity is 0 or 13m/s
 
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If the balloon is ascending at 13m/s, then vi=13m/s (with the sign convention that upwards is positive). Try using the equation y=vi*t+0.5a*t^2.
 
thanks, got part a right.

for b, I'm getting 42.18. is that right and would it be negative if it is going down?
used vf= vi + a*t and vf^2=vi^2+2*a*d
 
suxatphysix said:
thanks, got part a right.

for b, I'm getting 42.18. is that right and would it be negative if it is going down?
used vf= vi + a*t and vf^2=vi^2+2*a*d

Yes and yes; although you only need to use the second equation.
 
ty got it right
 

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