Hey,

May I ask what is a gravitational force? And how about acceleration du to gravity, I understand it is the attraction or something from Earth with a constan of 9.18 m / 2s

OK. So what dos per secon per secon

The gravity of Earth, denoted g, refers to the acceleration that the Earth imparts to objects on or near its surface. In SI units this acceleration is measured in metres per second per second (in symbols, m/s2 or m·s-2) or equivalently in newtons per kilogram (N/kg or N·kg-1). It has an approximate value of 9.81 m/s2, which means that, ignoring the effects of air resistance, the speed of an object falling freely near the Earth's surface will increase by about 9.81 metres (about 32.2 ft) per second every second. This quantity is sometimes referred to informally as little g (in contrast, the gravitational constant G is referred to as big G)

is weight what? mass x gravity



????

 

It's 9.81m/\(s^2\) near the surface of the Earth.

Per second per second means that for each second you fall, your velocity increases 9.81 meters per second. It's 9.81m/s per second, or 9.81m per second per second. So after \(t\) seconds your velocity (if you started at velocity zero) would be \(9.81t\)m/\(s^2\).

Weight is the measure of a force exerted on an object by gravity. It's measured in newtons, and is mass x acceleration due to gravity.

 

Thanks for your answer, it gives me some insight.

So, would that mean that it would take 2 seconds for 29.43 meters?

Also, what's the mass x acceleration due to gravity. For example, I have 1 kilogram rice x 9.81 m/s2. Then what?

 

If you have a one kilogram mass sitting on a table in a gravitational field whose strength at that point is 9.81N/kg, then that mass will excert 9.81N of force on the table.

If that same mass were dropped (in a vacuum) from that same point it would accelerate at (9.81m/s)/s.

Applying a little calculus, you would find that the total distance, d, travelled by an object in time t, starting from rest and accelerating at a constant rate, a, would be:

d = 0.5a(t^2)

So after 2 seconds, the distance travelled would be

d = a*(2 sec^2)

If a is 9.81m/s^2, then this would be a distance of 19.62m.

The thing you have to remember is that, at the end of 1s, the object is moving at a velocity of 9.81m/s, it has not travelled 9.81m in that time because, for most of that time, it was going slower. In fact, it's average speed is just half of the speed at the end of the 1s.

This is one way to get that equation above. An object starting from rest and accelerating uniformly reaches a speed after time t of a*t. It's average speed during that same inverval is half of this, or Vavg=0.5at. It travelled at that averate speed for a time t, so the distance travelled is d=Vavg*t=0.5at*t=0.5at^2.

 

This is Physics 101. First semester college. The very best way to undertand is to suffer through doing the math in a first semester text book. It is the work of doing the math that makes the logic of the acceleration and accumulated velocity come together in your head.

 

Many non-physics people giggle when they hear "per second per second"
simply because they have no clue what they are talking about.

The expression "per second per second" is totally erroneous.

You mean to say "(metres per second) per second"

which translates to "velocity per second"

which translates to acceleration.

In mathematical terms,

velocity v = Δx/Δt, where x = distance and t = time

acceleration = Δv/Δt

Δ means change or difference. Hence Δx means the change in position x and Δt means the time it takes to travel Δx.

In dimensional analysis, we say this is in units of distance/time/time.

Hence the unit of acceleration is \(ms^{-2}\)

 

Just so you folks know, Lightfire is just starting high school. Adjust your explinations accordingly.