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

  • #1

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.
 

Answers and Replies

  • #2
773
0
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?
 
  • #3
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.
 
  • #4
1,065
10
Taking down as positive.
[itex]\ddot {y}=-b\dot y+g[/itex]
or
dv/dt=-bv+g
This is a differential equation of motion.
 
Last edited:
  • #5
Taking down as positive.
[itex]\ddot {y}=-b\dot y+g[/itex]
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.
 
  • #6
1,065
10
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.
 
  • #7
I am sure that there is another way to do this because we have not learned about SUVAT or differential equations.
 
  • #8
773
0
The time it takes for an object to fall from a height H is [itex]\sqrt{\frac{2h}{g}}[/itex]

It should be easy to set two sides of an equation equal to each other (moon and earth) and solve for h.
 
  • #9
The time it takes for an object to fall from a height H is [itex]\sqrt{\frac{2h}{g}}[/itex]

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
[itex]\sqrt{\frac{2h}{g}}[/itex] = [itex]\sqrt{\frac{2h}{g}}[/itex]

then won't h cancel out? also, you haven't taken into account the air resistance on earth?
 
  • #10
773
0
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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