Physics Projectile Word Problem -- Dropping packages from an airplane

AI Thread Summary
The discussion revolves around calculating the distance a pilot must release packages from a C-17 aircraft to ensure they land on target while traveling at a speed of 263 m/s from an altitude of 3048 m. The time for the packages to reach the ground is calculated to be approximately 25 seconds, leading to a horizontal distance of 6556 meters that the aircraft must be ahead of the drop zone. Participants emphasize the importance of clearly defining variables and visualizing the problem through diagrams. Misunderstandings arise regarding the interpretation of elapsed time and the relationship between the aircraft's position and the drop zone. Ultimately, the correct distance for the package drop is confirmed to be 6556 meters ahead of the target.
Retweetfire
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Homework Statement


US Air Force C-17 Globemaster III aircraft were used to airdrop food to Afghan refugees in the early stages of the war against the Taliban in 2001. Calculate how far ahead of the drop zone a pilot would have needed to drop humanitarian aid packages if the delivery occurred at a speed of 263 m/s and an altitude of 3048 m to avoid enemy fire. Assume the packages were released from their containers immediately and neglect air resistance.

Homework Equations


t = 2H / g
D = vt

The Attempt at a Solution


It takes 24.928 s for the package to reach the floor.
 
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Retweetfire said:
the distance the aircraft will have at this time, which is 6556.064 m/s.
m/s? That's not a distance.
Anyway, what do you mean by "distance the aircraft will have at this time"? The time you calculated is an elapsed time, not a point in time. Which point in time do you mean, and distance of the aircraft from where?
 
haruspex said:
m/s? That's not a distance.
Anyway, what do you mean by "distance the aircraft will have at this time"? The time you calculated is an elapsed time, not a point in time. Which point in time do you mean, and distance of the aircraft from where?
*6556.064
You are right its a distance based on an elapsed time, but I can't think of another way to find the position when the delivery occurred.
 
Retweetfire said:

Homework Equations


t = 2H / g
D = vt

You'd want to define your variables more clearly. Additionally, drawing a picture of what the problem asks for will always help in these types of problems.

##\vec v_0=\langle263,0\rangle \frac{m}{s}##
##h=3048 m##
 
Retweetfire said:
*6556.064
You are right its a distance based on an elapsed time, but I can't think of another way to find the position when the delivery occurred.
It is the right thing to do, but I'm struggling to understand where your difficulty lies.
In the context of the problem, what distance do you think this represents? How is that different from the distance you need to find?
 
haruspex said:
It is the right thing to do, but I'm struggling to understand where your difficulty lies.
In the context of the problem, what distance do you think this represents? How is that different from the distance you need to find?
My difficulty lies, in what things I must find in order to get the initial position of the aircraft.
 
Retweetfire said:
My difficulty lies, in what things I must find in order to get the initial position of the aircraft.
You don't need the exact position, just the distance from the drop zone.
Please try to answer my other question. When you multiplied the horizontal speed of 263m/s by the descent time of 25 seconds, what distance did you think that would calculate? (And why did you later delete that from post #1?)
 
haruspex said:
You don't need the exact position, just the distance from the drop zone.
Please try to answer my other question. When you multiplied the horizontal speed of 263m/s by the descent time of 25 seconds, what distance did you think that would calculate? (And why did you later delete that from post #1?)
1. that would calculate the range after delivery.
2. I deleted it because I thought it was irrelevant to the solution.
 
Retweetfire said:
that would calculate the range after delivery.
By range, do you mean the horizontal travel of the parcel or of the aircraft?
By "after delivery" do you mean from releasing the parcel or from landing of the parcel?
 
  • #10
haruspex said:
By range, do you mean the horizontal travel of the parcel or of the aircraft?
By "after delivery" do you mean from releasing the parcel or from landing of the parcel?
1. The horizontal travel of the aircraft(but it can also apply to the parcel)
2. the releasing of the parcel.
 
  • #11
Retweetfire said:
1. The horizontal travel of the aircraft(but it can also apply to the parcel)
2. the releasing of the parcel.
Ok. So if that is the horizontal travel of the parcel after it is released, how far ahead of the target does it need to be released?
 
  • #12
haruspex said:
Ok. So if that is the horizontal travel of the parcel after it is released, how far ahead of the target does it need to be released?
I've hit a wall, I don't know.
 
  • #13
Retweetfire said:
I've hit a wall, I don't know.
As Eclair_de_XII suggested, draw a diagram. Show the horizontal movement of the aircraft and the trajectory of the parcel. Mark the horizontal distance you calculated for the parcel after release. How far from the landing point is the release?
 
  • #14
haruspex said:
As Eclair_de_XII suggested, draw a diagram. Show the horizontal movement of the aircraft and the trajectory of the parcel. Mark the horizontal distance you calculated for the parcel after release. How far from the landing point is the release?
Still can't get the answer, man I'm hopeless.
 
  • #15
Retweetfire said:
Still can't get the answer, man I'm hopeless.
Suppose you are throwing rocks at a target on the ground some distance away. You can throw the rock 20m. How far from the target do you need to be when you throw?
 
  • #16
haruspex said:
Suppose you are throwing rocks at a target on the ground some distance away. You can throw the rock 20m. How far from the target do you need to be when you throw?
20 m
 
  • #17
Retweetfire said:
20 m
Right. So if the parcel continues on for 6556m after it is released, how far does the plane need to be in front of the target when releasing it?
 
  • #18
haruspex said:
Right. So if the parcel continues on for 6556m after it is released, how far does the plane need to be in front of the target when releasing it?
yes, it is 6556. Man what a bummer. I'm just devastaded because this is from a problem from a higher edition Phyics book, and ,as I was given an answer to "this" problem number, I started searching for the explanation. But it results that the questions are different, that's why I had such a hard time.
 
  • #19
Retweetfire said:
yes, it is 6556. Man what a bummer. I'm just devastaded because this is from a problem from a higher edition Phyics book, and ,as I was given an answer to "this" problem number, I started searching for the explanation. But it results that the questions are different, that's why I had such a hard time.
Not sure I follow. Are you saying you had reason to believe that answer wrong? If so, it would have helped if you had said that at the start.
 
  • #20
haruspex said:
Not sure I follow. Are you saying you had reason to believe that answer wrong? If so, it would have helped if you had said that at the start.
Basically. Man I'm sorry I wasted your time. Sorry for any inconsistencies with my English - second language.
 
  • #21
Retweetfire said:
Basically. Man I'm sorry I wasted your time. Sorry for any inconsistencies with my English - second language.
ok, no problem.
 
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