Medical supplies dropped from a plane (projectile motion)

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SUMMARY

The discussion centers on the physics of projectile motion as it pertains to dropping medical supplies from a plane at an altitude of 150m and a horizontal speed of 40m/s. The calculations confirm that the supplies must be released 220m ahead of the campsite to ensure they land accurately. The time of descent is calculated to be approximately 5.5 seconds, which is derived from the equation Δdv = 1/2 aΔt². The pilot cannot wait until directly above the campsite because the supplies will land 220m away due to the horizontal velocity.

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  • Understanding of projectile motion principles
  • Familiarity with kinematic equations, specifically Δdv = 1/2 aΔt²
  • Knowledge of horizontal displacement calculations
  • Basic trigonometry for visualizing motion
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Humbleness
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Homework Statement


A plane is to be used to drop emergency medical supplies (encased in a crash-proof container) to campers in Northern Ontario. The campers are in a clearing. The plane will approach the clearing at an altitude of 150m while flying level at 40m/s.

a) Why can't the pilot wait until the plane is directly over the campsite before dropping the supplies?
b) How far (in meters) ahead of the campsite should the supplies be dropped? Justify your choice.

Homework Equations


Δdv= 1/2 aΔt2 - To find how long until the supplies hit the ground.
horizontal displacement = horizontal velocity which is 40m/s x Δt - To find how far horizontally would the supplies travel after being released from the plane

The Attempt at a Solution


I will first start with b)
So I used the formula:
Δdv = 1/2 aΔt2

150m [down] = 1/2 (9.8m/s2 [down]) Δt2
Δt2 = 150 : 4.9 = √30.6
Δt = 5.5sThen I used the horizontal velocity of 40m/s x 5.5s (the time), to get the answer to horizontal displacement:
40m/s x 5.5s = 220m

So the supplies should be dropped 220m ahead of the campsite. This means that both the plane and the supplies themselves will be at 220m ahead of the campsite at the same time. Am I correct with this part, at least? I would really appreciate a confirmation.

For a) I tried drawing a diagram to show the airplane directly above the campsite, and then tried to draw a triangle, so that perhaps using trigonometry I could find out why the pilot shouldn't drop the supplies when directly above the campsite. But I am still not understanding how to even begin the calculations, using what equation... My main concern is that I am not understanding how to get to the answer to part a. How do I calculate what would happen if the plane dropped the supplies when directly above the campsite? I don't understand.
 
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Hi Humbleness and welcome to PF.

Humbleness said:
How do I calculate what would happen if the plane dropped the supplies when directly above the campsite? I don't understand.
You have already calculated that in part (b). When the package is released, it will land 220 m horizontally away from the drop point whether there is a campsite directly below or not. This, of course, assumes your calculation of 220 m is correct.
 
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Thank you kuruman for the welcome and quick response! I really appreciate it.
So then, my calculations answered both a and b? I am confused in that case as to how I would vindicate my response to question a. What exactly is it asking of me?
 
Humbleness said:

Homework Statement

Then I used the horizontal velocity of 40m/s x 5.5s (the time), to get the answer to horizontal displacement:
40m/s x 5.5s = 220m

So the supplies should be dropped 220m ahead of the campsite. This means that both the plane and the supplies themselves will be at 220m ahead of the campsite at the same time. Am I correct with this part, at least? I would really appreciate a confirmation.

Yes, this is correct, and hence question (a) is answered :) This shows that if you wait for the stuff to be dropped directly above, the supplies will land 220m away from the campsite
 
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Understood clearly now. Thank you so much for helping me clarify the confusion!
 
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