How to Calculate the Height of a Falling Object Using Basic Physics Equations?

[papa_smurf493]

Homework Statement

Hi I am just starting a college physics class, all the work is pretty much online and I am am having a terrible time learning all the equations and things of that nature. The question is

You are standing by your living room window when an object falls vertically downward in front of you. You observe that it takes 0.210 seconds for the object to pass the window, which is 1.35 m tall. From what height did the object fall from above the window?


I know that the object is experiencing 9.8 m/s in freefall but i have no clue where to start as far as equations go. Please help I am terrible at this.

Homework Equations

The Attempt at a Solution

 

[papa_smurf493]

Im sorry that i can't really provide any work, I am definitely not trying to freeload, and i do want to learn how to do this stuff. I just need someone to walk me through it.

 

[LowlyPion]

Welcome to PF.

You just need to know that the distance an object falls is given by:

x = 1/2*g*t²

Since you know the difference in time and you know the distance it traveled in that additional time then you just subtract one equation for one distance from the other. And since 1 time is longer by the time they give you ... then just solve the resulting quadratic equation.

 

[papa_smurf493]

what is the g in the equation? I am guessing x would be the height and t is the time

 

[fwFAWFSERG]

im not too sure, what i would probably work towards would be:
well it took the object .210 seconds to move 1.35 meters, so find out that initial velocity (when it first is visible from the window)

then you can use the velocity you solved for (when the object is at the top of the window/ first visible) and use that as the final velocity again, and solve for the distance.

edit: g is gravity, and lowlys method is good too.

 

[The Dagda]

Already answered.

 

[papa_smurf493]

so my equation should look like this

x=1/2 * 9.8m/s * .210 Squared

 

[papa_smurf493]

V= 9.8m/s Squared *.210 gives me 2.058m/s squared that's using the forumula V=gt is this correct for the initial velocity?

 

[papa_smurf493]

still need some help please

 

[LowlyPion]

so my equation should look like this

x=1/2 * 9.8m/s * .210 Squared

No. The .210 is the time difference.

X1 = 1/2*9.8*t1²
X2 = 1/2*9.8*t2²

And you know that t2 = t1 + .210
And x2 - x1 = 1.35

The rest is algebra.

 

[papa_smurf493]

im still lost sorry i must just be stupid

 

[papa_smurf493]

how do i go about finding what t1 is?

 

[LowlyPion]

Sorry, I won't solve it for you.

It won't do either of us any good.

Just keep trying.

 

[papa_smurf493]

ok then can you explain it in a different way

i have no idead how to find t1

 

[papa_smurf493]

ok here's what I am trying to think of in my head. Basically the question states that it takes a object .210seconds to cover 1.35meters. So with that given information i have to find what height the object falls from. Well a given is the objects acceleration is 9.8m/s squared

 

[papa_smurf493]

ok i may be stupid or something, this is suppose to be easy and i ve been working on it for like a hour please help

 

[papa_smurf493]

ok i cam up with this. Will this equation work?

D=Vi *t +1/2 * a * t2

0*.210+1/2*9.8*?

 

[papa_smurf493]

am i that lost that nobody can help me?

 

[papa_smurf493]

i guess it would help if i found the avg speed which is 1.35/.210 = 6.43m/s

 

[papa_smurf493]

AVG Speed = 1.35/.210 = 6.43 m/s
a = (vm - vt)/t
9.81 = (6.43 - vt)/(0.105)

vt = 5.40 m/s

5.402 = 0 2 + 2(9.81)d


d = 1.49 m
Object was released 1.49 m above window.

is this correct?

 

[papa_smurf493]

wow this is a great forum, i really appreciate all the replys I am getting and the help! You guys are great!

 

[LowlyPion]

I already supplied you with a solution if you had bothered to solve it.

https://www.physicsforums.com/showpost.php?p=2052023&postcount=10

I'm uncertain what more I can do.

 

[papa_smurf493]

well you could have broke it down a little better and explained what i needed to do instead of thowing a formula at me and sayin use it

 

[Kaizer6]

Any basic mechanics book solves this problem within the first couple of chapters, take a book outfrom the library on classical mechanics and do some reading. You'll understand what youre trying to do then.

 

[The Dagda]

https://www.physicsforums.com/showthread.php?t=287752

This book is excellent and free and covers all you need to know about the basics, Courtesy of Motionmountain via Greg Bernhardt.

Follow the link:

http://www.motionmountain.net/

I downloaded it yesterday as I'm a little rusty on this stuff, and it is perfect for pre-degree studies.

z(t)=z_0+v_0(t-t_0)-\frac{1}{2}g(t-t_0)^2

z(t) = height
z_0 = initial height
t_0 = time fall starts
g = 9.8m/s²

Probably too late for this to help but meh. For more information check out page 74 of that text.

 

[papa_smurf493]

see the thing that got me was the time fall starts, i didnt know if i plugged in the time threw the window or solved for it. Because i was also looking for height

 

[The Dagda]

see the thing that got me was the time fall starts, i didnt know if i plugged in the time threw the window or solved for it. Because i was also looking for height

The problem is we're not allowed to just give you the answer, so all we can do is give you a push in the right direction, obviously once you'd handed it in it doesn't matter any more. Check out the rules. What happened with your homework?

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[AEM]

Homework Statement

Hi I am just starting a college physics class, all the work is pretty much online and I am am having a terrible time learning all the equations and things of that nature. The question is

The Attempt at a Solution

papa_smurf493, Check your private messages

 

[papa_smurf493]

ya i did read that, i just needed someone to explain it in away that makes sense to me. I went to my lab today and i am obviously strugglin with this subject so I am going to a study group that meets every week. I really do appreciate your guy's help, i just am not the kind of person to figure out equations when there thrown at me. You have to explain which each element is to me. The reason this homework was such a rush was because it was given the day it was due. It was suppose to be an easy problem but it turned out to be a pain for me. I just passed my analytical algebra and trig class last semester with flying colors, but for some reason this class is giving me trouble. We do a lot of stuff on masteringphysics.com and i hate it.

 

[The Dagda]

I have the derivation of the equation s=ut+1/2at^2 if you want me to run through it, it's the same thing Newton did when he invented calculus hundreds of years ago. It's actually just the integration of the equations mentioned earlier. Integration is the area under a graph, so it is the area where speed is related to time. Since speed is also related to distance traveled by acceleration due to gravity we can insert some values in, and rearrange the equation to give us the equation above. Try that textbook as well it goes into detail how the equations relate to the real world.

People often have a strength in pure over applied maths and vise a versa.

In everyday life, animals and humans agree on the Euclidean properties of velocity,
space and time. In particular, this implies that a trajectory can be described by specifying
three numbers, three coordinates (x, y, z) – one for each dimension – as continuous
functions of time t. (Functions are defined in detail on page 826.) This is usually written
as x = x(t) = (x(t), y(t), z(t)). For example, already Galileo found, using stopwatchand ruler, that the height z of any thrown or falling stone changes as

<br /> z(t)=z_0+v_0(t-t_0)-\frac{1}{2}g(t-t_0)^2<br />

where t0 is the time the fall starts, z0 is the initial height, v0 is the initial velocity in the
vertical direction and д = 9.8m/s2 is a constant that is found to be the same, within about
one part in 300, for all falling bodies on all points of the surface of the Earth. Where do
the value 9.8m/s2 and its slight variations come from? A preliminary answer will be
given shortly, but the complete elucidation will occupy us during the larger part of this
hike.
Equation (5) allows us to determine the depth of a well, given the time a stone takes
to reach its bottom.The equation also gives the speed v with which one hits the ground
after jumping from a tree, namely v =\sqrt{2gh} . A height of 3myields a velocity of 27 km/h.
The velocity is thus proportional only to the square root of the height. Does this mean
that one’s strong fear of falling results from an overestimation of its actual effects?
Galileo was the first to state an important result about free fall: the motions in the
horizontal and vertical directions are independent. He showed that the time it takes for
a cannon ball that is shot exactly horizontally to fall is independent of the strength of the
gunpowder, as shown in Figure 41.Many great thinkers did not agree with this statement
even after his death: in 1658 the Academia del Cimento even organized an experiment
to check this assertion, by comparing the flying cannon ball with one that simply fell
vertically. Can you imagine how they checked the simultaneity? Figure 41 also shows
how you can check this at home. In this experiment, whatever the spring load of the
cannon, the two bodies will always collide in mid-air (if the table is high enough), thus
proving the assertion.
In other words, a flying canon ball is not accelerated in the horizontal direction. Its
horizontal motion is simply unchanging. By extending the description of equation (5)

Two expressions for the horizontal coordinates x and y, namely:

x(t) = x0 + vx0(t − t0)
y(t) = y0 + vy0(t − t0), (6)

A complete description for the path followed by thrown stones results.A path of this shape
is called a parabola; it is shown in Figures 18, 41 and 42. (A parabolic Page 45 shape is also used
for light reflectors inside pocket lamps or car headlights. Can you show why?)
Physicists enjoy generalizing the idea of a path. A path is a trace left
in a diagram by a moving object. Depending on what diagram is used, these paths have
different names. Hodographs are used in weather forecasting. Space-time diagrams are
useful to make the theory of relativity accessible. The configuration space is spanned by
the coordinates of all particles of a system. For many particles, it is has a high number of
dimensions. It plays an important role in self-organization.The difference between chaos
and order can be described as a difference in the properties of paths in configuration
space. The phase space diagram is also called state space diagram. It plays an essential
role in thermodynamics.