From what height will two objects hit the ground at the same speed?

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Homework Help Overview

The problem involves two objects of equal mass dropped from a height, one on Earth experiencing air resistance and the other on the Moon without air resistance. The goal is to determine the height from which both objects will hit the ground at the same speed, considering the different gravitational accelerations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to derive equations for the time of fall for each object as a function of height and question how to equate them.
  • Some mention using differential equations to model motion with air resistance, while others express uncertainty about the variables involved.
  • There is mention of using SUVAT equations for the Moon's motion, but some participants indicate a lack of familiarity with these concepts.
  • Concerns are raised about the cancellation of height in equations and the impact of air resistance on the Earth.

Discussion Status

The discussion is ongoing with various approaches being explored, including differential equations and energy methods. Some participants are questioning their understanding of the problem and the relevance of certain equations, indicating a productive exchange of ideas without a clear consensus yet.

Contextual Notes

Participants note that they have not yet learned about certain mathematical tools like SUVAT or differential equations, which may limit their ability to solve the problem effectively.

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Homework Statement



There are 2 objects both with mass 0.50 kg which are dropped from rest at height h. One is dropped on Earth and experiences air resistance and one is dropped on moon and experiences no air resistance. The gravitational acceleration on Earth is 9.8 and on moon it is 1.6.

From what height h will the objects hit the ground at the same speed?

Homework Equations


F=-kv for air resistance
k=0.17

The Attempt at a Solution



I know that we will somehow use the gravitational acceleration.
 
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Well I'd start by figuring out the equations that will tell you how long it will take for each object to fall (it'll be a function of h). Then I'd set the two equal to each other.

Do you see where you would get those equations?
 
Vorde said:
Well I'd start by figuring out the equations that will tell you how long it will take for each object to fall (it'll be a function of h). Then I'd set the two equal to each other.

Do you see where you would get those equations?

Unfortunately, no I don't see where to get the equations because if I set it equal the h will cancel out.
 
Taking down as positive.
\ddot {y}=-b\dot y+g
or
dv/dt=-bv+g
This is a differential equation of motion.
 
Last edited:
azizlwl said:
Taking down as positive.
\ddot {y}=-b\dot y+g
or
dv/dt=-bv+g
This is a differential equation of motion.

Hello,

I am looking for at what height h will the velocity of the two objects be constant. I don't see how this while help solve. Also, I don't know what each of the variables stands for.
 
You can use SUVAT equation for the height of the motion on the moon.
With air resistance of the earth, its a bit complicated to find the height with differential equation involved.
 
I am sure that there is another way to do this because we have not learned about SUVAT or differential equations.
 
The time it takes for an object to fall from a height H is \sqrt{\frac{2h}{g}}

It should be easy to set two sides of an equation equal to each other (moon and earth) and solve for h.
 
Vorde said:
The time it takes for an object to fall from a height H is \sqrt{\frac{2h}{g}}

It should be easy to set two sides of an equation equal to each other (moon and earth) and solve for h.

Im not sure I understand because when I make
\sqrt{\frac{2h}{g}} = \sqrt{\frac{2h}{g}}

then won't h cancel out? also, you haven't taken into account the air resistance on earth?
 
  • #10
Oh I totally misread the original question, I'm sorry for misleading you with my stuff. I thought wanted something different.

I can think of two ways I'm pretty sure I could solve this with (Energy stuff, and calculus stuff), but neither factors in air resistance, so I'll need to think about it for a tiny bit.

Again, sorry for talking about the wrong stuff.
 

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