Projectile Motion - Distance between Packages

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SUMMARY

The discussion centers on the physics of projectile motion, specifically the distance between two packages dropped from a plane. The correct answer to the question posed is that the distance between the packages increases over time due to gravitational acceleration. The first package falls 4.9 meters after one second, while the second package, dropped one second later, falls 4.9 meters after its own first second, resulting in an increasing distance of 14.7 meters after two seconds. The analysis confirms that the horizontal component of velocity remains constant while the vertical distance increases.

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion
  • Familiarity with gravitational acceleration (9.8 m/s²)
  • Knowledge of kinematic equations, specifically Sv=1/2 at²
  • Ability to analyze motion in both horizontal and vertical components
NEXT STEPS
  • Study the principles of projectile motion in detail
  • Learn about kinematic equations and their applications in real-world scenarios
  • Explore the effects of air resistance on projectile motion
  • Investigate the concept of relative motion in physics
USEFUL FOR

Students studying physics, educators teaching projectile motion, and anyone interested in understanding the dynamics of falling objects in a vacuum.

MohammadG
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Homework Statement


A package of supplies is dropped from a plane and one second later a second package is dropped. Neglecting air resistance, the distance between the falling packages will:

A) Be constant
B) decrease
C) increase
D) depend on their weight

Homework Equations


Sv=1/2 av t^2

The Attempt at a Solution


Well I thought it would be constant but my teacher informed me it was C. He didn't explain why though. I'm assuming the planes speed is constant, which means both package will be subject to the same horizontal component of velocity. However the first package would have already fallen (4.9 meters) downwards due to gravitational acceleration. The second one would then be released and would be subject to the same forces. (As horizontal component of velocity would remain constant)

So I don't see how they would have an increasing distance.

My idea was,

Sv=1/2(9.8)(1)^2 = 4.9

Sv=1/2(9.8)(2)^2 = 19.6

Sv=1/2(9.8)(3)^2 = 44.1


Therefore, when package 1 is at 4.9, package 2 is at 0 (displacement). [difference 4.9 m]
When package 1, is at 19.6, package 2 is at 4.9. [difference of 14.7m]

etc and its increasing.

I think this is the right answer. How can I explain this in a sentence using physics terms. Its a multiple choice question so no calculations were needed even though I did some.

Thanks.
 
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Which direction, x-or -y is the question referring?
 
physicsisgrea said:
Which direction, x-or -y is the question referring?

Both x and y. However as the horizontal component won't change, we can just use the vertical component correct?

As both have the same trajectory.
 

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