Motion of a projectile thrown from a plane

AI Thread Summary
A helicopter flying horizontally at 8.5 m/s at an altitude of 18 m ejects a package backward at 10 m/s relative to the helicopter. To determine the horizontal distance between the package and the helicopter when the package hits the ground, the time to fall can be calculated using T=sqrt(2h/g), resulting in approximately 1.9 seconds. The package's horizontal speed relative to the ground is -1.5 m/s, while the helicopter continues moving forward at 8.5 m/s. The total horizontal distance covered by the package is calculated to be 19 meters when accounting for both speeds. The discussion highlights the importance of correctly understanding relative motion in projectile problems.
drewbie55
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Homework Statement


A helicopter is flying horizontally at 8.5 m/s and an altitude of 18 m when a package of emergency medical supplies is ejected horizontally backward with a speed of 10 m/s relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?


Homework Equations


Not quite sure.


The Attempt at a Solution


None. I'm thinking I have to find time first but I'm not sure how to do that either.
 
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To find time, it would be T=sqrt(2h/g) right. I did that then did d=vt for both the helicopter and the package thrown and added them together but still didn't get the right answer.

Any suggestions?
 
Why did you add them? The package is ejected backwards. (The easy way is to find and use the horizontal speed of the package with respect to the ground.)
 
Sounds right. I get t=1.9sec to reach ground
Package is moving backward at (10-8.5) m/s the helicopter is moving forward at 8.5 m/s
so total distance is (8.5+10-8.5)*1.9 = 19m

You can simply pick the helicopter as the frame of reference, then it is just thrown out at 10m/s for 1.9s
 
I added the two distances because I thought that would give the total distance.
 
Thanks. I see what I did wrong and why I wasn't getting the right answer.
 
drewbie55 said:
I added the two distances because I thought that would give the total distance.
What if the package were ejected at a speed of 8.5 m/s backwards with respect to the plane? What would the net horizontal distance traveled be then?
 
As I said, it's a bit of tricky question ( or a good intro to relativity)
Because it is thrown backward tou have to subtract the forward speed of the helicopter but if you are measuring from the groudn you then add the speed of the helicopter back in - they cancel. Consider viewing it from the helicopter (or imagine the helicopter is hovering) then it is just thrown backward with 10m/s.
 
mgb_phys said:
Package is moving backward at (10-8.5) m/s the helicopter is moving forward at 8.5 m/s
so total distance is (8.5+10-8.5)*1.9 = 19m
You are miscalculating the speed. The package is moving at -10 m/s relative to the plane; the plane is moving at +8.5 m/s relative to the ground. The speed of the package relative to the ground is -10 +8.5 = -1.5 m/s backwards.
 
  • #10
Yes but the helicopter is still moving 8.5m/s forward (relative to ground) after the drop so the separation on the ground is increasing at 10m/s
 
  • #11
mgb_phys said:
Yes but the helicopter is still moving 8.5m/s forward (relative to ground) after the drop so the separation on the ground is increasing at 10m/s
D'oh! I misread the question. :redface: Sorry about that! You are absolutely correct.
 
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