Packaged dropped from Hot Air Balloon

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Homework Help Overview

The problem involves a package being dropped from a hot air balloon that is ascending at a constant speed. The scenario is set at a height of 19 meters above the ground, and participants are tasked with determining the time the package remains in the air after being released, considering gravitational acceleration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to solve for time, questioning the definitions of variables such as displacement, initial velocity, and acceleration. There is confusion regarding the direction of the initial velocity relative to the direction of acceleration due to gravity.

Discussion Status

Some guidance has been provided regarding the application of the kinematic equation, but participants are still exploring the implications of the initial upward velocity and the downward displacement. Multiple interpretations of the variables are being discussed, and there is no explicit consensus on how to proceed with the calculations.

Contextual Notes

Participants are grappling with the sign conventions for upward and downward motion, particularly how to represent the initial velocity and displacement in the context of the problem. There is also a focus on ensuring the correct application of the kinematic equation given the scenario's conditions.

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Homework Statement


A Hot air balloon is traveling vertically up-
ward at a constant speed of 2.6 m/s. When
it is 19 m above the ground, a package is
released from the balloon.
The acceleration of gravity is 9.8 m/s^2.
After it is released, for how long is the
package in the air? Answer in units of s.


Homework Equations


http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L6a1.gif

The Attempt at a Solution


I tried doing X = .5at^2 + vit and using -2.6, but it didnt work.
 
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you can solve the problem directly with:

d = v1*t + (1/2)at^2

what is d? what is v1? what is a?
 
well a is -9.8 because of gravity.

Vi is 2.6, but its the opposite direction of falling. So what do I do with it?

d is 19 m.

I don't know how to use that equation because the Velocity is upward, but the object is falling DOWN.
 
dvdqnoc said:
well a is -9.8 because of gravity.

Vi is 2.6, but its the opposite direction of falling. So what do I do with it?

d is 19 m.

I don't know how to use that equation because the Velocity is upward, but the object is falling DOWN.

d is -19m. because the object drops a distance of 19m. ie: vertical displacement = final height - initial height = 0m - 19m = -19m.

plug in a = -9.8, vi=2.6 and d = -19 into the equation...

solve for t.
 

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