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Free fall Experiment Help

by bilalbajwa
Tags: experiment, fall, free
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bilalbajwa
#1
May28-07, 06:30 PM
P: 28
1. The problem statement, all variables and given/known data

Experiment:-
OBJECTIVES
In this experiment you will investigate the motion of a freely falling body on a fictitious planet.
You will discover for an object that falls from rest
(1) the value of the acceleration due to gravity on the planet,
(2) how the distance fallen depends on the elapsed time, and
(3) an interesting relation between the instantaneous velocity at the center of a time interval and the average velocity over that interval.

SECTION 1 http://docs.google.com/Doc?id=dctgg53g_53f3p8zf
SECTION 2 http://docs.google.com/Doc?id=dctgg53g_57qn8qzw


3. The attempt at a solution
Guys i have already done this experiment and submitted it. But though i have no idea what i did and what are the conclusions.
GUYS PLEASE EXPLAIN THESE EXPERIMENTS TO ME> I SINCERILY SAY THAT I DID THESE EXPERIMENTS AND IT TOOK ME ALOT OF TIME. SO I DID PUT ALOT OF EFFORT IN THEM. BUT SAD THAT THEY ARE NOT MAKING ABY SENSE TO ME>
Please help Me

Please give me the concept so that i can answer these questions!!!
QUESTIONS
Q1 Explain any difference you find between the values in Table I, column 6 and Table I, column 8.
Q2 Based on your data in Table I, how does the displacement in free fall depend on the time elapsed since the object was dropped?
Q3 Based on your data in Table II, what is the relation between the average velocity for a given time interval and the instantaneous velocity in the interval?
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dontdisturbmycircles
#2
May28-07, 06:36 PM
dontdisturbmycircles's Avatar
P: 560
For #1 use the fact that acceleration is the rate at which velocity is changing. Since it is a constant acceleration, can you use a bit of math to find out how much the velocity increases in m/s in any given second?

edit: woops sorry bilalbajwa I assumed that the 'objectives' were the questions, my bad. Let me look at the tables...
bilalbajwa
#3
May28-07, 06:39 PM
P: 28
Q1.
Average Acceleration = delta x/ delta ok i understand
but what is yi/ti^2???

WHY yi/ti^2 is half of avg acceleration???

WHats the logic comming out of it???

dontdisturbmycircles
#4
May28-07, 06:59 PM
dontdisturbmycircles's Avatar
P: 560
Free fall Experiment Help

The yi/ti^2 is the vertical displacement divided by the time interval squared.

Basically what is happening is when they are spitting out the 'velocity' they are measuring the instantaneous velocity! When you are looking at yi/ti you are looking at the AVERAGE velocity over that specific time interval. I will suppy some pictures in a second to help.
dontdisturbmycircles
#5
May28-07, 07:04 PM
dontdisturbmycircles's Avatar
P: 560
Actually I don't think you need pictures to understand this, I have to go shopping as well(will be back soon). But plot the average velocities yi/ti for each interval on the same graph as the recorded instantaneous velocities, what do you notice about the slope of each?
dontdisturbmycircles
#6
May28-07, 07:10 PM
dontdisturbmycircles's Avatar
P: 560
I am trying to write quick while making this as clear as possible. What did I show you in the other thread? You can use the midpoint rule for a line to derive an equation for the average velocity when you are dealing with constant acceleration. It is 1/2(vi+vf) right? So basically at the end of the experiment, if you look at the average velocity, it is only half of the actual final velocity and if you use that to compute the acceleration, you end up with 1/2 the actual acceleration, get it?
bilalbajwa
#7
May28-07, 07:29 PM
P: 28
OK one thing i got here:
SECTION I
Coloumn comparison "Velocity" vs yi/ti
"Velocity" = Instantaneous Velocity
yi/ti = Average Velocity

Conclusion: In a free falling body, when we take equal time intervals and unequal displacements. Avg Velocity is half of Instantaneous velocity???
AM I RIGHT

IF YESY WHY IS IT SO???
dontdisturbmycircles
#8
May28-07, 10:59 PM
dontdisturbmycircles's Avatar
P: 560
It is due to the fact that there is a constant acceleration, not specifically because it is a 'free falling body'. Can you tell me why avg velocity is 1/2 of the 'instantaneous velocity at the END OF THE INTERVAL.'?
bilalbajwa
#9
May28-07, 11:28 PM
P: 28
Because at the end of the interval Instantaneous velocity is on the peak and AVG velocity is the averages of the inital + final. (i am not sure)
dontdisturbmycircles
#10
May28-07, 11:47 PM
dontdisturbmycircles's Avatar
P: 560
I think you are getting there and probably understand. But anyways, the AVERAGE velocity of an object over a specific time interval (under constant acceleration) is

[tex]v_{ave}=\frac{v_{i}+v_{f}}{2}[/tex]
Since Vi is 0 in your the case of your object, you could express

[tex]v_{ave}=\frac{v_{f}}{2}[/tex]

Another way of expressing [tex]v_{ave}[/tex] is

[tex]v_{ave}=\frac{\Delta d}{\Delta t}[/tex]

Where delta d is the change in displacement and delta t is the change in time

So this is where the yi/ti comes from, its simply the average velocity.

I feel that I am not doing a very good job of explaining this.. :/


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