Free Fall Experiment (1)

8/10/2019 Free Fall Experiment (1)

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Trinity University of Asia

St. Lukes College of Nursing

FREE FALLExperiment _____

SN: 8, 18, 20, 42Cabral, Nolan

Ceralde, Keith BryanMarino, Kristia

Nones, Kevin

February 17, 2014


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I. Objective:

This experiment is being conducted to determine the gravitational force acting

on an object and its final velocity when it is being dropped from determined

distance (height).

II. Theory:

Free fall is described as the motion of an object which is ONLY under the

influence of gravity (g) of earth without air resistance. The gravitational force on

a body or an object in a state of free fall is uniform and is constant. The

direction of the gravitational force accelerates downwards and is directed

towards the center of the earth and its value is 32 ft/s2or 9.8 m/s2or 980 cm/s2

at sea level.

III.

Apparatus:

(1)Golf Ball (1) Marble Meter Stick

Set-up:

Materials:

(1) Golf Ball

(1) Marble

Meter stick

Stop watch

Note: This experiment is not limited to any object being trialed.

Have enough space where these objects may be dropped.


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IV.

Procedure

After all materials have been gathered, select a member of the group to measure and

mark on the wall 1 meter (m), 1.5 m, and 2.0m. Once that has been done, another

member will grab one of the trial objects, golf ball or marble, and drop it from a certain

height; another member should measure the time of when the object was released to

when it hits the ground. Time is measured in seconds. Once data has been gathered

record and log into data table.

Note: Initial velocity of an object being dropped is 0 m/s

Voyand to

dy(ht.)

Vy and tf

Formulas needed to calculate gravitational force and final velocity

Gravity: g

Distance: d

Time: s

Initial Velocity: Voy

Final Velocity: Vfy

V=d/t

g= Vfy-Voy/ t

d = (Voy)(t) + g(t)2

2gd= (Vfy)2(V0y)

2

% difference: actual gravityexperimental gravity

Actual gravity


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V.

Data/Results/Computation :

Marble (heavy ball)

Distance Time Vfy Distance Time Vfy Distance Time Vfy

Trial 1

Trial 2

Trail 3

1.0 m

1.0 m

1.0 m

0.48 s

0.48 s

0.40 s

2.08 m/s

2.08 m/s

2.50 m/s

1.5 m

1.5 m

1.5 m

0.53 s

0.67 s

0.49 s

2.83 m/s

2.23 m/s

3.06 m/s

2.0 m

2.0 m

2.0 m

0.77 s

0.72 s

0.64 s

2.60 m/s

2.78 m/s

3.12 m/s

Mean 1.0 m 0.45 s 2.22 m/s 1.5 m 0.56 s 2.68 m/s 2.0 m 0.71 s 2.82 m/s

Gravitational Force:

at distance of 1.0 m: at distance of 1.5 m: at distance of 2.0 m:

g= 2.22 m/s0 m/s g= 2.68 m/s0 m/s g= 2.82 m/s0 m/s

0.45 s 0.56 s 0.71 s

g= 4.93 m/s2 g= 4.78 m/s2 g= 3.97 m/s2

Final Velocity:

2(4.93m/s2)(1.0m)= Vfy2-(0)2 2(4.78 m/s2)(1.5m) = Vfy

2-(0)2 2(3.97 m/s2)(2.0m) = Vfy2-(0)2

2(4.93m/s2)(1.0m)= Vfy2 2(4.78 m/s2)(1.5m) = Vfy2 2(3.97 m/s2)(2.0m) = Vfy23.14 m/s = Vfy 3.79 m/s = Vfy 3.98 m/s = Vfy

% Difference:

9.8 m/s24.93 m/s2 9.8 m/s24.78 m/s2 9.8 m/s23.97 m/s2

9.8 m/s2 9.8 m/s2 9.8 m/s2

% diff. = 0.496 % diff. = 0.512 % diff. = 0.595

Golf ball (light ball)

Distance Time Vfy Distance Time Vfy Distance Time Vfy

Trial 1

Trial 2Trail 3

1.0 m

1.0 m1.0 m

0.46 s

0.48 s0.44 s

2.17 m/s

2.08 m/s2.27 m/s

1.5 m

1.5 m1.5 m

0.66 s

0.56 s0.48 s

2.27 m/s

2.68 m/s3.12 m/s

2.0 m

2.0 m2.0 m

0.75 s

0.83 s0.93 s

2.66 m/s

2.41 m/s2.15 m/s

Mean 1.0 m 0.46 s 2.17 m/s 1.5 m 0.56 s 2.68 m/s 2.0 m 0.84 s 2.41 m/s

Gravitational Force:

at distance of 1.0 m: at distance of 1.5 m: at distance of 2.0 m:

g= 2.17 m/s0 m/s g= 2.68 m/s0 m/s g= 2.41 m/s0 m/s

0.46 s 0.56 s 0.84 s

g= 4.71 m/s2 g= 4.78 m/s2 g= 3.91 m/s2

Final Velocity:

2(4.71 m/s2)(1.0 m) = Vfy2-(0)2 2(4.78 m/s2)(1.5m) = Vfy

2-(0)2 2(3.91 m/s2)(2.0 m) = Vfy2-(0)2

2(4.71 m/s2)(1.0 m) = Vfy2 2(4.78 m/s2)(1.5m) = Vfy2 2(3.91 m/s2)(2.0 m) = Vfy2

3.07 m/s = Vfy 3.79 m/s = Vfy 3.95 m/s = Vfy

% Difference:

9.8 m/s24.71 m/s2 9.8 m/s24.78 m/s2 9.8 m/s23.91 m/s2

9.8 m/s2 9.8 m/s2 9.8 m/s2

% diff. = 0.519 % diff. = 0.512 % diff. = 0.601


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VI.

Analysis of Data:

After all of the data had been gathered within this experiment our

group has concluded that the acceleration of an object is similar to one

another as it falls within a certain height or distance.

Acceleration of an object from its height is the difference of final

velocity and initial velocity divided by time(g= Vfy-Voy/ t). The

experimental values of acceleration (gravitational force) on the objects

resulted in similar values l ike when it was dropped at a height of 1.5

meters. The gravitational force is 4.78 m/s2on both experimental

objects even though each object had a different mass. The other heights

as well resulted in nearly similar values showing that they accelerate

downwards at the same speed and time. The final velocities of ball one (marble) compared to ball two (golf ball)

was just hundredths of a fraction off , showing that the speed of decent

is closely similar to one another.

On the other hand, our percentage difference is 51% to 60% from the

actual gravitational force which indicates experimental variables such as

air resistance, accuracy of time recording on release, and stop of object.

These variables played a large role in making such a big difference in

experimental gravitational force and actual gravitational force since free

fall is the motion of objects that are fall ing under the sole influence of

gravity.


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VII.

Conclusion:

In conclusion with this experiment the gravitational force was similar to each object

even though each object did not have the same mass. The variables such as air

resistance and timing affected the experimental values and percentages greatly

indicating that these factors played a major difference in the experiments outcome.

Also, the gravitational force or acceleration of an object is only under free fall when

there are no other forces acting upon it but the force of gravity. The experimental

results indicated a close similarity in values showing that each objects accelerated at

almost the same rate but with the gravitational force being almost 50% to 60% of the

actual gravitational force.

Newtons 2ndLaw is the net force of an object is equal to its mass times acceleration (F=

m x a). Acceleration due to gravity is then interpreted as (a= F/m). This example is to

show that if the two objects are only under the force of gravity without air resistance

they will accelerate at the same rate regardless of mass.

(10g= 0.01 kg) (1g= 0.001 kg)

Golf ball Marble

F(gravity)

Fg= 0.098N Fg= 0.0098N

a= Fnet/mass

a= 0.098 N/ .01 kg a= 0.0098N/ .001kg

a= 9.8 m/s2 a= 9.8 m/s2


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VIII.

Problem Sets:

A rock is dropped 80 meters from a cliff. How long does it take to reach the

ground?

Vo

2gdy=Vf2 - Vo2

2(-9.8m/s2)(-80m)= Vf2(0)

1568 m2/s2 = Vf2

dy=80 m a=g 39.60 m/s = Vf

g= VfV0/t ; t = V fVo/g

t= 39.60 m/s0 m/s

9.8 m/s/s

Vf t= 4.04 seconds.

An object is thrown straight upward with an initial speed of 8 m/s and strikes the

ground 3 seconds later. What is the maximum height the ball reached?

C

Vf = 0 m/s Total time= AC + CB

2gdy= Vf2V0

2

2(-9.8 m/s2)dy= 0 (8 m/s)2

dy = - 64 m2/s2

2(-9.8 m/s2)

dy= 3.26 m

V0= 8 m/s

A B

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Free Fall Experiment (1)