Free Fall Acceleration Problem

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Homework Help Overview

The problem involves a scientific balloon ascending while an instrument package free-falls after breaking free. The discussion centers on determining the maximum height the package reaches after detaching and its height above the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to find the height above the break-free point and questions the correctness of their calculations for the second part of the problem.
  • Some participants analyze the timing of the package's ascent and descent, referencing a graph to identify when the package returns to the break-free point.

Discussion Status

Participants are exploring different aspects of the problem, including the timing of the package's motion and the implications of its velocity at specific times. There is an ongoing examination of the original poster's calculations and interpretations.

Contextual Notes

The original poster has provided specific values for initial velocity, time, and acceleration, but there may be uncertainties regarding the application of kinematic equations and the interpretation of the graph referenced in the discussion.

chounx93
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Homework Statement



As a runaway scientific balloon ascends at 19.6 m/s, one of its instrument packages breaks free of a harness and free-falls. The figure below gives the vertical velocity of the package versus time, from before it breaks free to when it reaches the ground.
2-p-053.gif


(a) What maximum height above the break-free point does it rise?
(b) How high is the break-free point above the ground?


Homework Equations





The Attempt at a Solution



i know part A is 19.6 m
i just can't seem to get the correct answer for part B

these are the variables that i have and i think i'll need
V0=0 m/s
t=6.0 s
a=-9.8 m/s/s
y-y0=?

i used this kinematic equation
y-y0=vt-1/2at^2
but the answer still isn't correct
 
Physics news on Phys.org
The package is at the break-free point at time t = 2 s and it is moving up. Can you figure out by looking at the graph at what later time it is at the same point, moving down?
 
it's at the same point at t=4 s
 
At t = 4 s the velocity is instantaneously zero. This mean maximum height.
 

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