Question

is anyone up to help me...

Answer 1

hey im on don't know how much i'd be able to help but i'll try

Answer 2

thanks! I have a quiz on Free fall in my General Physics class tomorrow and this whole course as been overwhelming..:/

Answer 3

okay well what do you need help with?

Answer 4

oh man..not sure where to start but here's a question from my homework.

A brick is released with no initial speed from the roof of a building and strikes the ground in 2.60s , encountering no appreciable air drag.


How tall, in meters, is the building?

Answer 5

A= 9.8, V initial= 0 and T=2.6...is this correct so far in what values I need'?

Answer 6

so i take it its free fall with no wind resistance or any external force?

Answer 7

correct

Answer 8

the values are correct im just flipping through which formulas you are stipulated to use.

Answer 9

alright

Answer 10

so what i can gather so far the formula which i would use is
s=ut+1/2*9.8*t^2

s=distance
u=initial velocity
t=time
g=9.8

all you do is plug your values into the formula and you should get a distance in meters.

Answer 11

so it should look something like this

s=0*2.6 + 1/2 *9.8 *2.6^2
s=0+4.9*6.76
s=33.124m

Answer=the building is 33.124m in height.

Answer 12

you need any other help?

Answer 13

one moment sorry I went to the bathroom

Answer 14

lol no problem

Answer 15

can you show me the formula you used please

Answer 16

\[s =ut +\frac{ 1 }{ 2}g \times t ^{2}\]

Answer 17

What does the u stand for?

Answer 18

initial velocity

Answer 19

Can you provide me with all the formulas used for free fall? This is what I have:

Y= Yi + Vi Delta(t) + 1/2ay X delta(t)^2

Vyf= Vyi + aDelta(t)

Vyf^2=Vyi + 2a + Deltay

Answer 20

wow that's confusing lol my formulas are simpler to understand. same thing though but different way of writing them.

Answer 21

I believe that's where my problem is

Answer 22

1: \[v ^{2}=u ^{2}+2gs\]
2: \[s=ut+\frac{ 1 }{ 2 }g t ^{2}\]
3: \[v=u+g t\]

okay just remember now though when you use these formulas please dont forget when you get a question with no gravitational force. but you get an acceleration the same formulas are used. when you use these formulas with an acceleration then you take g and replace with a for acceleration.

Answer 23

so are those the same formulas that I gave you but just simplified and easier to look at?

Answer 24

yeah pretty much. the ones i gave you are more appealing to the eye.

Answer 25

alright so here's the next part of the same question

How fast is the brick moving just before it reaches the ground?

Answer 26

try the formulas i gave you and see if they work for you
oh here is the description list of my formulas
s=distance
v=final velocity
u=initial velocity
t=time
g=gravitational acceleration
a=acceleration

Answer 27

is the gravitational acceleration the same as constant acceleration?

Answer 28

okay so if you are given an acceleration of an object that is not the same as the gravitational force then you will not use 9.8 you will have to use whats given. but in majority of the time if we are given a free falling object we use gravitational force.

lets say we are given an object that is forced upward with an acceleration of 4.5m/s then we use the given acceleration until the object starts to free fall in which case we use the gravitational force.

try the question you asked me about the velocity of the brick and see what you get. i already have the answer so i can cross check with your answer.

Answer 29

alright

Answer 30

25.5 m/s...?

Answer 31

I did the third formula

Answer 32

I GOT IT RIGHT! Yes lol

Answer 33

are you still there...

Answer 34

hey im here sorry got disconnected
yeah 25.5m/s is correct lol congrats

Answer 35

alright so I'm stuck...

Answer 36

A tennis ball on Mars, where the acceleration due to gravity is 0.379 of a g and air resistance is negligible, is hit directly upward and returns to the same level 6.10s later.


How high above its original point did the ball go?

Answer 37

haha with?

Answer 38

okay so this is what i was talking about when they give you a different acceleration. now take the new value 0,379 as g so g= 0.379 t=6.10 and initial velocity was not stated so the object as an initial velocity of 0. u=0

Answer 39

okay I'm getting the wrong answer, this is what I'm doing for the equation::
s=0*6.10 + 1/2 *.379 *6.10^2

Answer 40

what answer do you get?

Answer 41

7.05

Answer 42

I'm thinking it's beacuse you use the wrong acceleration. If I understand you correctly, the acceleration should be 0.379 of g (9.8). So 0.379*9.8=3.7142 m/s^2 ;)

Answer 43

And also you're using the wrong time! It takes 6.10 s for the ball to come back down? In free fall that means it takes half that time going to the top of its trajectory ;)

Answer 44

.... got 1.72 when i switched out the numbers and still got it wrong... :/

Answer 45

3.05 for the time right?

Answer 46

s=0*3.05 + 1/2 *3.7142 *3.05^2

Answer 47

Yeah, that should give you the right answer :P

Answer 48

ill try again

Answer 49

unfortunately it's incorrect...unless I'm doing my math wrong...

Answer 50

Well, what do you get?

Answer 51

so can you post the question again but a little better than the first time

Answer 52

1.72 again

Answer 53

I get 17.2 :P

Answer 54

yeah i get 17.2 aswell

Answer 55

I'm crazy. Idk what I did wrong

Answer 56

but you guys got it correct

Answer 57

I'm gonna go rest my eyes for 30 mins ill be back

Answer 58

Yeah, I was about to suggest that! ;)

Answer 59

enjoy the rest