Airplane drops a package (relative motion)

In summary, a plane flying at a speed of 275 m/s at an altitude of 3000 m drops a package, which takes approximately 24.74 seconds to hit the ground. The package will hit the ground directly under the point of release due to the absence of wind resistance. The position of the plane at the time of impact can be calculated by multiplying its speed (
  • #1
southernbelle
35
0

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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  • #2
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?
 
  • #3
I did

Yf = Yi + Via + 1/2at2

0= 3000 + 0 + -4.9t2

t=25 seconds?
 
  • #4
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:
  • #5
Ok, so
0=3000 + 0 + -4.9t^2

And I got 7.5 seconds?
 
  • #6
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?
 
  • #7
Okay, thanks.
But how did you get

3000 divided by 9.8t2?

What formula did you plug those numbers in to?
 
  • #8
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
 
  • #9
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
 

1. How does the height of the airplane affect the speed of the dropped package?

The height of the airplane has no direct effect on the speed of the dropped package. The speed of the package is determined by the initial velocity of the airplane and the force of gravity acting on the package.

2. How does the angle at which the package is dropped affect its trajectory?

The angle at which the package is dropped can greatly affect its trajectory. If the package is dropped at a horizontal angle, it will have a longer horizontal distance to travel before hitting the ground. If the package is dropped at a vertical angle, it will hit the ground in a shorter amount of time but with a greater vertical velocity.

3. How does air resistance affect the motion of the dropped package?

Air resistance, or drag, can greatly impact the motion of a dropped package. As the package falls, it will experience air resistance which will slow it down. The amount of air resistance depends on the size and shape of the package, as well as the density and speed of the air through which it is falling.

4. Does the mass of the package affect its acceleration when dropped from an airplane?

The mass of the package does not affect its acceleration when dropped from an airplane. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Since the force of gravity is the same on all objects, the mass of the package will not affect its acceleration.

5. Can the position and speed of the airplane affect the landing spot of the dropped package?

Yes, the position and speed of the airplane can affect the landing spot of the dropped package. If the airplane is moving at a high speed, the package will have a greater horizontal velocity when it is dropped, causing it to land further away from the airplane. Additionally, the position of the airplane at the time of the drop will also determine the landing spot of the package.

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