How to caclulate the acceleration of a dropped object.

[mdmaaz]

I know the formula for calculating the gravitational attraction between two objects. Suppose there is a planet, and I know the gravitational attraction it will have on the object. The object is not very far from the surface of the planet. If I drop the object, how can calculate the acceleration and time it will take for the object to reach the surface.

 

[JaredJames]

It's acceleration will be g.

If you know the g value, you know the acceleration.

 

[mdmaaz]

I know the a value of g, but I don't know the acceleration, please explain.

 

[KingNothing]

Force = mass * acceleration.

If you know the force and the mass of the object, you can find acceleration.

 

[Bloodthunder]

Force on dropped object = object's weight
F=W
mg = ma
g = a

 

[JaredJames]

Neglecting air resistance, all objects of negligible mass in comparison to the planet will accelerate at the same rate towards the ground (hammer and feather on the moon ring any bells?).

If you include air resistance, things get a bit more complicated.

F=ma is not the right way to go about this.

 

[BruceW]

If you ignore air resistance (i.e. this only works for short distances for fairly dense objects), then just use g as the acceleration, and then integrate twice to get:
x = \frac{1}{2} g t^2 + v_0 t + x_0
So if you're dropping the object a height h above earth, then v_0 is zero and x - x_0 is h, so you get:
h = \frac{1}{2} g t^2
and then rearrange to get the time taken t.

Of course for things like feathers, air resistance is important even for a small height. But for things like rocks and people, you can ignore air resistance for a small height. (i.e. you can correctly calculate how much time it takes a person to land after jumping off a diving board).

 

[mdmaaz]

Is "acceleration" the same thing as the "speed" of the object?

 

[mdmaaz]

What does the "t" stand for in BruceW's explanation?

 

[BruceW]

t means time after the object was dropped. so when t=0 the object is at the position it was dropped from, and for t greater than zero, it will have fallen some distance.
acceleration is the rate of change of speed, and speed is the rate of change of distance.
You should probably get familiar with basic calculus to be able to better understand the physics of this problem. An introductory textbook to maths and/or physics would be useful.
In this problem, the acceleration due to gravity is roughly constant at the Earth's surface, and the speed is gt (assuming the object was dropped with no initial speed). This means that at a given time t (for example 2 seconds), then the speed will equal g times 2 seconds = roughly 20 meters per second. (The speed is downwards, since that is the direction of the acceleration).

 

[mdmaaz]

Thanks BruceW for helping me so much. I'm just a thirteen year old trying to learn physics.
And what does "v" and "x" stand for?

 

[JaredJames]

v is velocity, x is distance.

You may want to try some simpler problems to kick off with. Perhaps take a look at the equations of motion here: http://en.wikipedia.org/wiki/Equations_of_motion#Classic_version and then try out some simple examples to become familiar with the terminology.

 

[BruceW]

No problem. the x stands for position, and v stands for speed. They are both functions of time, and v is what you get when you differentiate x.
In my equation above, v_0 and x_0 mean the speed and position when t = 0. So they are simply constants.
I'm guessing you've done F=ma in school? As soon as you know a, then you can calculate speed and position as a function of time, by using integration.
The simplest situation is when a is constant, so they'll probably teach you that first.

 

[mdmaaz]

What is "g"? How do you calculate the "g" (gravitational constant)? Does the value of "g" change from place to place? And what is "a"

 

[douglis]

mdmaaz...read carefully the link that JaredJames gave you and if you still have questions we will be glad to answer them.

 

[JaredJames]

What is "g"? How do you calculate the "g" (gravitational constant)? Does the value of "g" change from place to place? And what is "a"

mdaaz, I agree with douglis, read what I posted and have a go with that. You're way above your head here and I think you should start with the basics.

G is the gravitational constant - it doesn't change, otherwise it wouldn't be a constant would it.

g is gravity.

They are not the same thing.

 

[russ_watters]

g is acceleration due to gravity.

 

[mdmaaz]

I read the link that JaredJames shared with me. It says that s stands for "displacement". What does "displacement" mean?

 

[JaredJames]

I read the link that JaredJames shared with me. It says that s stands for "displacement". What does "displacement" mean?

Google it.

By the sound of it you are right at the beginning, you need to start from basics.

You are clearly trying to dive into the deep end and unfortunately, without the basic concepts you aren't going to get very far.

 

[BruceW]

With this kind of stuff, the best way to learn it is when your teacher explains it and gets the class to do examples. But I guess there's nothing wrong with finding out a bit on your own :)

 

[mdmaaz]

I guess Jared James is right. Perhaps I should start with the basics, but what are the basics. I would love to learn this at school but I live in India and schools here usually don't teach these things. But I'm really passionate about physics which is why I'm trying to learn about it on my own.

 

[BruceW]

The bbc bitesize website is pretty good I think.
It goes KS1, then KS2, then KS3 then finally GCSE. (In order of how you should learn it).
That's the structure of secondary schools in England. The bitesize website is for students who want to revise online, and it pretty much teaches it as well.

 

[BruceW]

I've just had a look at it myself, and it doesn't seem to have a huge amount on physics, but it does have a lot on maths. And to understand physics, you generally need to have some maths skills, so its a good place to start.