# Weights and Speed?

Hi,
I am new to physics and the physics forum, I have a question.

I have a series of weights, 2kgs, 4kgs, 6kgs etc, held 10 meters above the ground.

How would I work out the speed and therefore time it takes for the weight to fall to the ground?

Also, how would I work out the acceleration?

If you could give me a detailed answer for just one of the weights that would be great.

Thanks
Dominic

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Danger
Gold Member
Welcome to PF, Dragonetti.
Your question is incomplete. The rate at which the weights drop, in atmosphere, depends upon how that weight is distributed. That's why parachutes work.
Gravitational acceleration at sea level on Earth is 32 ft/sec2. That, again, assumes no noticeable atmospheric drag.

Take the equasion : x=Vi + 1/2(gt^2).
X = distance
Vi = initial velocity
G = gravitational acceleration (10 m/s^2)
T = time

The mass of the weights will have no effect on the velocity with which the weight hits the ground.
Your initial velocity will be 0 bacause the weight will be let loose from a stationary position, 10 meters above the ground.
Take the time from the moment you let the weight loose 'till it hits the ground.

The velocity with which the weight hits the ground will be obtained by Vf^2=Vi^2 + 2gx.

On earth the estimated acceleration is 9.8 meters per second squared. Weight is not a factor when calculating acceleration and velocity, only resistance being air pressure for example. If you were to have a feather in a vacuumed tube it would fall the same speed as a brick of gold.
Hope that helps.
TM

Welcome to PF, Dragonetti.
Your question is incomplete. The rate at which the weights drop, in atmosphere, depends upon how that weight is distributed. That's why parachutes work.
Gravitational acceleration at sea level on Earth is 32 ft/sec2. That, again, assumes no noticeable atmospheric drag.
Hi thanks,
The weights are solid metal.

Mentallic
Homework Helper
Hi thanks,
The weights are solid metal.
Assuming they're spherical and not a sheet of metal or some other weird shape, it's safe to assume from that height and for those weights, they will all drop at the same time. If you notice any difference in drop times, more than likely it was the experimenter's inaccuracies than the atmospheric drag.

Take the equasion : x=Vi + 1/2(gt^2).
X = distance
Vi = initial velocity
G = gravitational acceleration (10 m/s^2)
T = time
In some countries, it is a federal offense to confound capital and minuscule letters representing physical constants or variables (not really, but it can easily bring about confusion).

HallsofIvy
Homework Helper
In some countries, it is a federal offense to confound capital and minuscule letters representing physical constants or variables (not really, but it can easily bring about confusion).
Punished by putting in the stocks!

I always laugh at medicine bottles that claim some number of MG in a pill. An Mg is a metric ton, right?

There's a 't' missing after the Vi in 'Take the equasion : x=Vi + 1/2(gt^2).'

Forgive my mistakes. I typed that post from my cell phone and can't use the lower case at the beginning of each sentence.
As for the forgotten t after Vi...that is just unforgivable, sorry.

Take the equasion : x=Vi + 1/2(gt^2).
X = distance
Vi = initial velocity
G = gravitational acceleration (10 m/s^2)
T = time

The mass of the weights will have no effect on the velocity with which the weight hits the ground.
Your initial velocity will be 0 bacause the weight will be let loose from a stationary position, 10 meters above the ground.
Take the time from the moment you let the weight loose 'till it hits the ground.

The velocity with which the weight hits the ground will be obtained by Vf^2=Vi^2 + 2gx.
Hi Frannas,