does anyone now how to work out the acceleration due to gravity with a free fall experiment???

Question

i no you use s=ut+0.5at(squared) but am not sure wat i do with that equation :P lol

Answer 1

You will need a pair of light gates, set vertically apart of s.

Ideally, this will not be so far apart that air resistance will play any part and to ensure you can not miss the second gate. They need to be far enough apart to allow accurate timing of the interval between the gates. Experiment

You will need to drop an object through. A rectangular piece of aluminium will do. Drop it as you would a blade, lengthways.
You will need to program your data logger to measure two speeds (u) and (v) from two gates and a time interval between them (t). The data logger will need to know the length of your object

The instantaneous speed will be greater through the lower gate than the top one.

the acceleration due to gravity (g) can be determined using the equation

v^2=u^2+2gs

rearrange to give

g= (v^2 - u^2)/2s

However, Your data logger can probably do all of that for you and just give you a numerical answer.

You would need to repeat several times and take an average.
There are errors involved in this which need taking account of when giving your answer.

Answer 2

That equation gives you the distance travelled in time t: If the initial velocity is 0, ut can be removed. Then s = 0.5 a t^2 where a is the acceleration and in this case is equal to g, the acceleration due to gravity.

Suppose you go to the top of a tower of height h and drop a small, smooth ball of steel or lead (so that you can ignore the air friction) and suppose you or your friend uses a sensitive stop watch to measure the time taken for the ball to hit the ground. Now, you have h, the height of the tower as s the distance travelled, g as the acceleration and t the time taken: h = 0.5 g . t^2 or g = 2h / t^2

I hope the working out is clear. I think Galileo's famous experiment from the leaning tower of Pisa had something to do with this type of reasoning.

Answer 3

A free-falling object is an object which is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. There are slight variations in this numerical value (to the second decimal place) which are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s/s in The Physics Classroom Tutorial in order to reduce the complexity of the many mathematical tasks which we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s/s, downward ( ~ 10 m/s/s, downward) Acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s/s means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then it would follow the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s/s.

Answer 4

If you are considering free fall from a short distance (i.e. close to sea level) then you can assume acceleration due to gravity is constant at 9.81 metres per second squared.

if you need more than that the the equation you have can help

s=ut+0.5at(squared)

s = distance traveled
u = initial velocity
t = time
a = acceleration

Answer 5

To work out the acceleration due to gravity, you must first re-arrange the equation s = ½at² to make a the subject thus:

s = ½at²
=
a = s/½t²
=
a = 2s/t²

1 ) Measure a drop in metres with as great a height as you can. This = s

2) Drop a pebble from the height & time the pebble hitting the deck in seconds, this = t

3) Square t to get t²

4) Work through the equation a = 2s/t² to get a

Answer 6

Use x = 0.5at^2, where x is the drop height.
Measure the free fall time t and solve for a = 2x/t^2.

Answer 7

s is the speed at which the object will travel.
A is the accelleration with which the object accellerates
(in a frictionless fall (on earth) this value is 9.81 meters per second per second (sometimes called the g instead of a))
t is the time in seconds.

im not sure what that 'ut' is doing there, but im assuming thats taking into account a startingspeed other than 0.

Answer 8

I seem to remember reading in the past somewhere that a free falling body accelerates at a rate of 32ft per second/per second. Hope this is of use and sorry it's imperial.

Answer 9

free fall means v(o)=0 and v= g t where g=9.81 MKS
. s(distance)=0.5 gt^2
.and the Limit velocity is mg =0.5 V^2 or V=(2 mg)rt constant.

Answer 10

The acceleration rate in freefall is 92 feet per second per second.