How high was the package above the ground when it was thrown

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Homework Help Overview

The discussion revolves around a physics problem related to kinematics, specifically the motion of a package thrown upward from a helicopter. The problem states that it takes 16 seconds for the package to strike the ground after being thrown at an initial velocity of 15 m/s, and participants are tasked with determining the height from which it was thrown.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial calculations regarding the time taken for the package to reach its maximum height and the distance covered during that time. There is confusion about whether the calculated distance represents the height of the helicopter or just the distance the package travels upward before stopping.

Discussion Status

Some participants have pointed out potential misunderstandings in the calculations, emphasizing the need to clarify the distinction between the height of the helicopter and the distance the package travels upward. There is acknowledgment of errors in reasoning, and one participant expresses gratitude for the guidance received.

Contextual Notes

Participants are operating under the assumption that air resistance can be ignored, and they are working with standard kinematic equations for vertical motion. The problem's parameters, such as the initial velocity and total time of flight, are central to the discussion.

Perseverence
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Homework Statement

( p 49 sterling)
Ignoring air resistance, if it takes 16 seconds for a package to strike the ground, how high above the ground was a package when it was thrown upward from the stationary helicopter at 15 m/s per second[/B]

Homework Equations


V=Vo+at
D=volt+(1/2)at^2

The Attempt at a Solution


It seems pretty straightforward that the solution would be the distance from the initial throw upward to the upward velocity becoming zero.
V=Vo+at
0=15+(-10)t
1.5 sec =time (to Vy=0)

Then plugging in 1.5sec into
D=volt+(1/2)at^2 to get 11.28m. But that is not the solution in the book. The solution in the book is 1040m.
 
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Perseverence said:

Homework Statement

( p 49 sterling)
Ignoring air resistance, if it takes 16 seconds for a package to strike the ground, how high above the ground was a package when it was thrown upward from the stationary helicopter at 15 m/s per second[/B]

Homework Equations


V=Vo+at
D=volt+(1/2)at^2

The Attempt at a Solution


It seems pretty straightforward that the solution would be the distance from the initial throw upward to the upward velocity becoming zero.
V=Vo+at
0=15+(-10)t
1.5 sec =time (to Vy=0)

Then plugging in 1.5sec into
D=volt+(1/2)at^2 to get 11.28m. But that is not the solution in the book. The solution in the book is 1040m.
You are making a massive mistake and missing steps. You are calculating how far up the book will go when its velocity becomes 0. That is 11.28.

However, the question asks for how high up the helicopter is, not how high the book went up before it became 0.

Try again.
 
lekh2003 said:
You are making a massive mistake and missing steps. You are calculating how far up the book will go when its velocity becomes 0. That is 11.28.

However, the question asks for how high up the helicopter is, not how high the book went up before it became 0.

Try again.
if you are in a helicopter and something in the air, when it reaches as high as it will go, it's Vertical Velocity will be zero. That is the distance from the helicopter
lekh2003 said:
You are making a massive mistake and missing steps. You are calculating how far up the book will go when its velocity becomes 0. That is 11.28.

However, the question asks for how high up the helicopter is, not how high the book went up before it became 0.

Try again.

Oh my gosh
You're right, I was calculating the wrong distance! I figured out what I was doing wrong. Thank you so much for your help!
 
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Perseverence said:
Oh my gosh
You're right, I was calculating the wrong distance! I figured out what I was doing wrong. Thank you so much for your help!
It's good that you realized. You're welcome.
 
The equation you are looking for is the kinematic equation for 1D vertical motion with no air drag. y = y0 + v0 t - gt^2 / 2 , t is the time of flight from y0 to y=0.
 

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