# Thread: Time taken for objects in free fall

1. ## Time taken for objects in free fall

Hi everyone.

Although energy equations can be used to prove the velocity of any mass in free fall, this there a method to prove the free fall of any mass in a time period.

by example; t = square root 2s/g but this method does not include either the energy of the mass, or the mass in question.

what I am therefore asking is does anyone know of a method to include the mass or energy into a equation or formula to find the time of any mass in free fall, neglecting air resistance.

many thanks

David

2. V = g t
V=dx/dt
=> dx = gt dt
Integrate both sides:

x = (1/2) g t^2

Is this what you want ?

-O

3. Originally Posted by oswaldo
V = g t
V=dx/dt
=> dx = gt dt
Integrate both sides:

x = (1/2) g t^2

Is this what you want ?

-O
Sorry that method does not include the energy of the masses involved in the free fall.

4. After falling a distance of s body of mass m loses potential energy equalt to $m * g * h$

Gain in kinetic velocity $1/2 m * v ^2
$

By law of conservation of evergy, loss in potential enrgy equals gain in kinetix energy
$=> m * g * h = 1/2 m * v^2$

Which gives $v = sqrt (2 g * h)$

Hope you are looking for the same.

Also if you assume that mass has initial velocity, the just use change of kinetic energy equal to change in potential energy. So this will introduce the quantity initial velocity in the equation.

5. Your sentense "Although energy equations ..." is misleading. I thought you don't want to use Energy conservation. Anyways, arpitagarwal82's approach is correct.

v=sqrt(2gx)
dx/dt = sqrt(2gx)
dx/sqrt(x) = sqrt(2g) dt

integrate both sides and you'll get:
x=0.5 gt^2

The same result.

-O

6. Originally Posted by arpitagarwal82
After falling a distance of s body of mass m loses potential energy equalt to $m * g * h$

Gain in kinetic velocity $1/2 m * v ^2
$

By law of conservation of evergy, loss in potential enrgy equals gain in kinetix energy
$=> m * g * h = 1/2 m * v^2$

Which gives $v = sqrt (2 g * h)$

Hope you are looking for the same.

Also if you assume that mass has initial velocity, the just use change of kinetic energy equal to change in potential energy. So this will introduce the quantity initial velocity in the equation.
Hi, I was not looking for the velocity of the mass in free fall, this i already worked out. What I was trying to establish was whether anyone new of any energy equation which would allow me to incorporate mass to actually prove the free fall velocity of any individual mass used.

by example, any mass in free fall neglecting air resistance could be found by t = sqrt 2s/g. But that equation does not include the mass, so could relate to just about anything if you see what i mean?

7. But free fall velocity doesn't depend upon mass. So how can you show any relation between them?

8. Originally Posted by arpitagarwal82
But free fall velocity doesn't depend upon mass. So how can you show any relation between them?
That is what I wanted to find, so if i said in a question that 2 tonnes of lead fell at 17.25 m/s, and then a feather also fell at 17.25 m/s, obviously without air resistance, then i just wanted to know if there was a method to include the masses so I could say these two items in free fall fell at the same rate using xyz equation or formula?

9. David, I think we're having some kind of communication problem here. Physical experiments show that objects in the vacuum -regardless of their mass- accelarates the same parameter towards center of the earth. That's what Newton observed and claimed. That is:

F=G m1 m2 / d^2
where
G is a universal constant
m1, m2 are masses,
d is the distance between masses.

Let m1 be the mass of Earth, and d is approx. it's radius, R, then:

F = m2 * a = G m1 m2 / R^2

m2's will cancel out. AND THAT'S WHY mass of the object has no effect. You'll have:

a = G m1 / R^2 all constants for Earth.
a here is called as g, and almost constant all over the world, about 9.8 m/s^2.

Objects regardless of their mass accelerates about 9.8 m/s for every passing second towards the center of Earth.

-O
No, Earth is not flat!

10. Originally Posted by oswaldo
David, I think we're having some kind of communication problem here. Physical experiments show that objects in the vacuum -regardless of their mass- accelarates the same parameter towards center of the earth. That's what Newton observed and claimed. That is:

F=G m1 m2 / d^2
where
G is a universal constant
m1, m2 are masses,
d is the distance between masses.

Let m1 be the mass of Earth, and d is approx. it's radius, R, then:

F = m2 * a = G m1 m2 / R^2

m2's will cancel out. AND THAT'S WHY mass of the object has no effect. You'll have:

a = G m1 / R^2 all constants for Earth.
a here is called as g, and almost constant all over the world, about 9.8 m/s^2.

Objects regardless of their mass accelerates about 9.8 m/s for every passing second towards the center of Earth.

-O
No, Earth is not flat!
Thank you for the information. In conclusion then;

The Earth is a Sphere, the acceleration due to gravity at the equator is 9.78 ms^2, and at other locations 9.81 ms^2.

The original question asked for a equation or formula to show any individual mass in free fall, it is therefore safe to say then that only controlled experiments would prove each mass in free fall, and there is no mathematical way of proof.