Physics Projectile Word Problem -- Dropping packages from an airplane

In summary: Still can't get the answer, man I'm...In summary, the 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.
  • #1
Retweetfire
12
0

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.
 
Last edited:
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  • #2
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?
 
  • #3
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.
 
  • #4
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##
 
  • #5
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?
 
  • #6
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.
 
  • #7
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?)
 
  • #8
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.
 
  • #9
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.
 

1. How do I calculate the initial velocity of a projectile dropped from an airplane?

To calculate the initial velocity, you can use the formula v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height from which the object is dropped.

2. How do I determine the time it takes for a package to hit the ground when dropped from an airplane?

The formula for calculating time is t = √(2h/g), where t is the time, h is the height, and g is the acceleration due to gravity. Plug in the known values and solve for t.

3. What factors affect the trajectory of a projectile dropped from an airplane?

The factors that affect the trajectory of a projectile are the initial velocity, angle of launch, air resistance, and the force of gravity. These factors can be manipulated to change the path of the projectile.

4. How do I account for air resistance when calculating the trajectory of a projectile?

To account for air resistance, you can use the formula F = 0.5ρAv², where F is the force of air resistance, ρ is the density of air, A is the cross-sectional area of the object, and v is the velocity. This force can be added to the calculations for trajectory.

5. Can I use the same equations for different units of measurement?

Yes, as long as the units are consistent, the same equations can be used for different units of measurement. For example, the acceleration due to gravity can be expressed in either meters per second squared or feet per second squared.

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