# Weight and acceleration due to gravity

 Page 3 / 3

1. Divide into pairs and explain Galileo's experiment to your friend.
2. Write down an aim and a hypothesis for Galileo's experiment.
3. Write down the result and conclusion for Galileo's experiment.

## Research project : experimental design

Design an experiment similar to the one done by Galileo to prove that the acceleration due to gravity of an object is independent of the object's mass. The investigation must be such that you can perform it at home or at school. Bring your apparatus to school and perform the experiment. Write it up and hand it in for assessment.

## Case study : determining the acceleration due to gravity 1

Study the set of photographs alongside showing the position of a ball being dropped from a height at constant time intervals. The distance of the ball from the starting point in each consecutive image is observed to be: ${x}_{1}=0$  cm, ${x}_{2}=4,9$  cm, ${x}_{3}=19,6$  cm, ${x}_{4}=44,1$  cm, ${x}_{5}=78,4$  cm and ${x}_{6}=122,5$  cm. Answer the following questions:

1. Determine the time between each picture if the frequency of the exposures were 10 Hz.
2. Calculate the velocity, ${v}_{2}$ , of the ball between positions 1 and 3.
${v}_{2}=\frac{{x}_{3}-{x}_{1}}{{t}_{3}-{t}_{1}}$
3. Calculate the velocity, ${v}_{5}$ , of the ball between positions 4 and 6.
${v}_{5}=\frac{{x}_{6}-{x}_{4}}{{t}_{6}-{t}_{4}}$
4. Calculate the acceleration the ball between positions 2 and 5.
$a=\frac{{v}_{5}-{v}_{2}}{{t}_{5}-{t}_{2}}$
5. Compare your answer to the value for the acceleration due to gravity ( $9,8\phantom{\rule{2pt}{0ex}}m·$ s ${}^{-2}$ ).

The acceleration due to gravity is constant. This means we can use the equations of motion under constant acceleration that we derived in  motion in one dimension to describe the motion of an object in free fall. The equations are repeated here for ease of use.

$\begin{array}{ccc}\hfill {v}_{i}& =& \mathrm{initial velocity}\left(\mathrm{m}·{\mathrm{s}}^{-1}\right)\mathrm{at}\phantom{\rule{2pt}{0ex}}\mathrm{t}=0\mathrm{s}\hfill \\ \hfill {v}_{f}& =& \mathrm{final velocity}\left(\mathrm{m}·{\mathrm{s}}^{-1}\right)\mathrm{at time}\phantom{\rule{2pt}{0ex}}\mathrm{t}\hfill \\ \hfill \Delta x& =& \mathrm{displacement}\left(\mathrm{m}\right)\hfill \\ \hfill t& =& \mathrm{time}\left(\mathrm{s}\right)\hfill \\ \hfill \Delta t& =& \mathrm{time interval}\left(\mathrm{s}\right)\hfill \\ \hfill g& =& \mathrm{acceleration}\left(\mathrm{m}·{\mathrm{s}}^{-2}\right)\hfill \end{array}$
${v}_{f}={v}_{i}+gt$
$\Delta x=\frac{\left({v}_{i}+{v}_{f}\right)}{2}t$
$\Delta x={v}_{i}t+\frac{1}{2}g{t}^{2}$
${v}_{f}^{2}={v}_{i}^{2}+2g\Delta x$

## Experiment : determining the acceleration due to gravity 2

Work in groups of at least two people.

Aim:  To determine the acceleration of an object in freefall.

Apparatus:  Large marble, two stopwatches, measuring tape.

Method:

1. Measure the height of a door, from the top of the door to the floor, exactly. Write down the measurement.
2. One person must hold the marble at the top of the door. Drop the marble to the floor at the same time as he/she starts the first stopwatch.
3. The second person watches the floor and starts his stopwatch when the marble hits the floor.
4. The two stopwatches are stopped together and the two times substracted. The difference in time will give the time taken for the marble to fall from the top of the door to the floor.
5. Design a table to show the results of your experiment. Choose appropriate headings and units.
6. Choose an appropriate equation of motion to calculate the acceleration of the marble. Remember that the marble starts from rest and that it's displacement was determined in the first step.
7. Write a conclusion for your investigation.
1. Why do you think two stopwatches were used in this investigation?
2. Compare the value for acceleration obtained in your investigation with the value of acceleration due to gravity ( $9,8\phantom{\rule{2pt}{0ex}}m·s{}^{-2}$ ). Explain your answer.

A ball is dropped from the balcony of a tall building. The balcony is $15\phantom{\rule{2pt}{0ex}}m$ above the ground. Assuming gravitational acceleration is $9,8\phantom{\rule{2pt}{0ex}}m·s{}^{-2}$ , find:

1. the time required for the ball to hit the ground, and
2. the velocity with which it hits the ground.
1. It always helps to understand the problem if we draw a picture like the one below:

2. We have these quantities:

$\begin{array}{ccc}\hfill \Delta x& =& 15\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \\ \hfill {v}_{i}& =& 0\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\hfill \\ \hfill g& =& 9,8\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-2}\hfill \end{array}$
3. Since the ball is falling, we choose down as positive. This means that the values for ${v}_{i}$ , $\Delta x$ and $a$ will be positive.

4. We can use [link] to find the time: $\Delta x={v}_{i}t+\frac{1}{2}g{t}^{2}$

5. $\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}t+\frac{1}{2}g{t}^{2}\hfill \\ \hfill 15& =& \left(0\right)t+\frac{1}{2}\left(9,8\right){\left(t\right)}^{2}\hfill \\ \hfill 15& =& 4,9\phantom{\rule{3.33333pt}{0ex}}{t}^{2}\hfill \\ \hfill {t}^{2}& =& 3,0612...\hfill \\ \hfill t& =& 1,7496...\hfill \\ \hfill t& =& 1,75\phantom{\rule{3.33333pt}{0ex}}s\hfill \end{array}$
6. Using [link] to find ${v}_{f}$ :

$\begin{array}{ccc}\hfill {v}_{f}& =& {v}_{i}+gt\hfill \\ \hfill {v}_{f}& =& 0+\left(9,8\right)\left(1,7496...\right)\hfill \\ \hfill {v}_{f}& =& 17,1464...\hfill \end{array}$

Remember to add the direction: ${v}_{f}=17,15\phantom{\rule{2pt}{0ex}}m·s{}^{-1}$ downwards.

By now you should have seen that free fall motion is just a special case of motion with constant acceleration, and we use the same equations as before. The only difference is that the value for the acceleration, $a$ , is always equal to the value of gravitational acceleration, $g$ . In the equations of motion we can replace $a$ with $g$ .

## Gravitational acceleration

1. A brick falls from the top of a $5\phantom{\rule{2pt}{0ex}}m$ high building. Calculate the velocity with which the brick reaches the ground. How long does it take the brick to reach the ground?
2. A stone is dropped from a window. It takes the stone $1,5\phantom{\rule{2pt}{0ex}}s$ to reach the ground. How high above the ground is the window?
3. An apple falls from a tree from a height of $1,8\phantom{\rule{2pt}{0ex}}m$ . What is the velocity of the apple when it reaches the ground?

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!