Airplane drops a package (relative motion)

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Homework Help Overview

The problem involves a plane flying at a speed of 275 m/s at an altitude of 3000 m dropping a package. The questions focus on determining where the package lands relative to both the point of release and the moving airplane, considering the absence of wind resistance.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the time it takes for the package to hit the ground and explore calculations involving vertical motion equations. Some express uncertainty about their results and seek clarification on the formulas used.

Discussion Status

There are multiple calculations and interpretations being explored regarding the time of fall and horizontal distance traveled by the package. Some participants provide guidance on how to approach the problem, while others question the accuracy of their calculations and the underlying assumptions.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion. There is a focus on understanding the physics of projectile motion without wind resistance.

southernbelle
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Homework Statement


A plane flying at a speed of 275 m/s at an altitude of 3000 m drops a package. The plane doesn't change its direction or speed and there is no wind resistance.
a) Where does the package hit the ground relative to the point of release?
b) Where does the package hit the ground relative to the moving airplane?


Homework Equations


R = sqrt A2 + B2


The Attempt at a Solution


a. Well I know that it would land directly under it because there is no wind resistance. But I don't know how to come up with that answer.

b. I did
R = 2752 +30002 and took the square root.
I came up with 3,013m.
 
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southernbelle said:

Homework Statement


A plane flying at a speed of 275 m/s at an altitude of 3000 m drops a package. The plane doesn't change its direction or speed and there is no wind resistance.
a) Where does the package hit the ground relative to the point of release?
b) Where does the package hit the ground relative to the moving airplane?


Homework Equations


R = sqrt A2 + B2


The Attempt at a Solution


a. Well I know that it would land directly under it because there is no wind resistance. But I don't know how to come up with that answer.

b. I did
R = 2752 +30002 and took the square root.
I came up with 3,013m.

How long did the package take to hit the ground from 3km?
 
I did

Yf = Yi + Via + 1/2at2

0= 3000 + 0 + -4.9t2

t=25 seconds?
 
Try;

Time for package to hit the ground.

3000m / -9.8m/s^2 = 306.12s^2 = 17.5s


Position of the plane at t = 17.5s, (time it takes for the package to hit the ground.)

17.5s x 275m/s = 4812.5m

Position of the package at t = 17.5s

Xf = Xi + Vi(t) + 1/2a(t)^2
Xf = 0 + 275/m/s(17.5s) + 0(17.5s)
Xf = 4812.5m

That's how I did it. Hope that helps!
 
Last edited:
Ok, so
0=3000 + 0 + -4.9t^2

And I got 7.5 seconds?
 
southernbelle said:
I did

Yf = Yi + Via + 1/2at2

0= 3000 + 0 + -4.9t2

t=25 seconds?

24.74 seconds, mostly close enough.

Now you know how fast it was going horizontally, so given the time and velocity, how far across the ground before the bomb hits? Horizontal speed times time perhaps?
 
Okay, thanks.
But how did you get

3000 divided by 9.8t2?

What formula did you plug those numbers in to?
 
LowlyPion said:
24.74 seconds, mostly close enough.

Now you know how fast it was going horizontally, so given the time and velocity, how far across the ground before the bomb hits? Horizontal speed times time perhaps?

So it would be 275m/s x 24.74s?
I got 6,803.5

That seems like such a big number and it ends up in m/s2
 
southernbelle said:
So it would be 275m/s x 24.74s?
I got 6,803.5

That seems like such a big number and it ends up in m/s2

6804 m.

Not the way I look at the units it's not m/s2

Now you have the altitude and the ground distance so you can calculate your displacement as you initially tried.

If the bomb hit on the ground at V*t where is the plane at this instant?

Edit: The number isn't all that far if you consider the speed is over 600 mph.
 
  • #10
southernbelle said:
Okay, thanks.
But how did you get

3000 divided by 9.8t2?

What formula did you plug those numbers in to?

The package fell 3000m at an acceleration of -9.8m/s^2 with a Vi of 0. Just divide the displacement (3000m) by a = -9,8m/s^2 and it give you the time it took the package to hit the ground in s^2. Then take the square root to obtain s.

d = a x t^2
3000m = -9.8m/s^2 x t^2
3000m / -9.8m/s^2 = t^2
306.12s^2 = t^2 = 17.5s
 
  • #11
LowlyPion said:
6804 m.

Not the way I look at the units it's not m/s2

Now you have the altitude and the ground distance so you can calculate your displacement as you initially tried.

If the bomb hit on the ground at V*t where is the plane at this instant?

Edit: The number isn't all that far if you consider the speed is over 600 mph.

Okay, thank you!
 
  • #12
eyvhgi557r said:
The package fell 3000m at an acceleration of -9.8m/s^2 with a Vi of 0. Just divide the displacement (3000m) by a = -9,8m/s^2 and it give you the time it took the package to hit the ground in s^2. Then take the square root to obtain s.

d = a x t^2
3000m = -9.8m/s^2 x t^2
3000m / -9.8m/s^2 = t^2
306.12s^2 = t^2 = 17.5s

D = 1/2 *a*t2
 

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