Solving for the Alaskan Rescue Team Equation

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An Alaskan rescue team drops a package from a plane flying horizontally at 40 m/s from a height of 100 m. The discussion focuses on calculating where the package will land and its velocity components just before impact. Participants express confusion over using the correct equations for vertical motion and the initial conditions, noting that the vertical velocity is zero at the moment of release. Clarification is provided that 100 m is the vertical distance and 40 m/s is the horizontal velocity. The conversation emphasizes the need to solve the equations for time of flight and displacement accurately.
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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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