Solving for the Alaskan Rescue Team Equation

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Homework Help Overview

The problem involves an Alaskan rescue team dropping a package from a plane traveling horizontally at 40 m/s from a height of 100 m. Participants are tasked with determining where the package lands relative to the drop point and the components of its velocity just before impact.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of free fall equations and express uncertainty about their applicability. There is confusion regarding the initial vertical and horizontal velocities, as well as the correct interpretation of the variables in the equations provided.

Discussion Status

Some participants are actively questioning the assumptions related to the initial velocities and the setup of the equations. Guidance has been offered regarding the interpretation of vertical and horizontal components, but no consensus has been reached on how to proceed with solving the equations.

Contextual Notes

Participants note that the problem is similar to other physics questions involving projectile motion, which may influence their approach. There is a lack of clarity on how to apply the equations correctly given the initial conditions.

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1. An Alaskan rescue team drops a package of emergency rations to a stranded hiker. The plane is traveling horizontally at 40 m/s at a height of 100m above the ground. A) Where does the package strik the ground relative to the point at which it was released? B) What are the horizontal and vertical components of the velocity of the package just before it hits the ground?


2. H=1/2gt^2
ΔX=ViTf+1/2ATf^2
AND apparently ΔY=ViTf+1/2ATf^2


3. So I'm not quite sure why you can't use the free fall equation...that was my first guess but it just seems too easy. The book says to use the third equation I put above, however I don't know how to completely solve it.
I got this far...
ΔY=ViTf+1/2ATf^2
100=?(would the Vi be 40m/s??)Tf+1/2(9.8)Tf^2
I know I'll end up with a Tf & Tf^2 on the right side of the equation, but I don't know what to do to solve.

Right, and then part B I'm soooo lost
 
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Let It Be said:
1. An Alaskan rescue team drops a package of emergency rations to a stranded hiker. The plane is traveling horizontally at 40 m/s at a height of 100m above the ground. A) Where does the package strik the ground relative to the point at which it was released? B) What are the horizontal and vertical components of the velocity of the package just before it hits the ground?2. H=1/2gt^2
ΔX=ViTf+1/2ATf^2
AND apparently ΔY=ViTf+1/2ATf^23. So I'm not quite sure why you can't use the free fall equation...that was my first guess but it just seems too easy. The book says to use the third equation I put above, however I don't know how to completely solve it.
I got this far...
ΔY=ViTf+1/2ATf^2
100=?(would the Vi be 40m/s??)Tf+1/2(9.8)Tf^2
I know I'll end up with a Tf & Tf^2 on the right side of the equation, but I don't know what to do to solve.

Right, and then part B I'm soooo lost


100m is vertical or horizontal?
40m/s is vertical or horizontal velocity?
PS: This ques is just like rolling marble ques.
 
cupid.callin said:
100m is vertical or horizontal?
40m/s is vertical or horizontal velocity?
PS: This ques is just like rolling marble ques.

100m is vertical
40m/s is horizontal
Could you help me figure out how to solve the ΔY=ViTf+1/2ATf^2 part please

P.P.S-Really? In that one I had to find Vi too...
 
Let It Be said:
100m is vertical
40m/s is horizontal
Could you help me figure out how to solve the ΔY=ViTf+1/2ATf^2 part please

P.P.S-Really? In that one I had to find Vi too...

here also you have it

initial horizontal velocity is same as of _____
and initial vertical is 0 as it is just dropped
 

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