General

@Imgr8haha: You're making me feel bad...I'm putting my time into research that uses tons of energy and probably leaves a sizeable footprint.

@ItzPootz: Maybe physical chemistry will make a good compromise, haha.

Also time for me to sleep so I don't die during tomorrow's classes.

@ItzPootz: I am thinking of doing a double major in chemistry + physics. chemistry major is like a prereq for the forensic program, and while im at uni, i'll try to do interns and stuff to get me into the field. But I might want to do criminal justice as well..so idk..you can only choose 2 majors no? o.o
I have to say, I am loving my quantum mechanics class. It's so fun..lol people thinks there's something wrong with me, that I'm so excited about class all the time.

@ItzPootz: I'm not all that crazy about "physics" physics...I guess I just like math. Btw, I did mention, or maybe I didn't that I want to be a detective when I grow up, because I like solving puzzles/mysteries/challenges..and of course, make this world a possibly "better" place.
But I do want to take more physics courses when I go to uni.
I'mma major in chemistry though.

</3 I need my beauty sleep now. Goodnight guys

OMG! I love that integral <3
Okay now I think there's something wrong with me..

That question was like this "pop-quiz" on my first day of quantum chem class D
Lol I'm also taking quantum mechanics at the same time
-not a cheater-

[quote=Oyster]What "higher level" course are you talking about
Lol me too, sometimes I usually look at it, and skip it if I see it's "straightforward"[/quote]
Well...maybe an example might be this thread.
http://www.basilmarket.com/forum/2510489/0/How_does_quantized_energy_fix_the_ultraviolet_catastrophe.html

I quite enjoyed it.

...even if the guy was trolling, I guess.

[quote=cb000]I suppose so. I guess I'm just feeling a slight desire to see someone seeking help for a slightly higher level course, since that would be refreshing and challenging to help with. And I really dislike plugging in specific values because I, like many other people, often get mixed up on the calculator.[/quote]
What "higher level" course are you talking about
Lol me too, sometimes I usually look at it, and skip it if I see it's "straightforward"

[quote=ItzPootz]@Oyster: http://imageshack.us/a/img827/6295/capturevvz.png I got the same thing

@cb000: well yeah, it's the same situation on Yahoo! Answers; only a very small percentage of students move past a general physics 1 course. Naturally you're much more likely to see (the same) kinematic questions over and over again.[/quote]
I suppose so. I guess I'm just feeling a slight desire to see someone seeking help for a slightly higher level course, since that would be refreshing and challenging to help with. And I really dislike plugging in specific values because I, like many other people, often get mixed up on the calculator.

[quote=Oyster]Lol there's a timer too?[/quote]

yea lol.
Everyone is busy doing their own thing.

Lol there's a timer too?
Maybe you should go online and chat with your friend while you're doing them.
Or do they generate different problems too? But same concept..I guess

[quote=Oyster]Yay two matching answers [/quote]

so the answer was 16.22. well the time was up. My score did go up thank you guys very much. Science is just not my thing. I might need help again later on T_T
Physic is needed for graduating.

[quote=Imgr8haha] 16.22 s[/quote]
Yay three* matching answers

I just got kicked off of basilmarket for some reason, when I refresh the page too much, or open another tab when I come back, my basil doesn't work.

[quote=Oyster]@nighttime: nooo...time up =/= time down
Okay time down is easier:
d=vi*t + 1/2 a*t^2
d= 297.133.....
a=+9.81m/s^2
vi=0
So we get t=sqrt{ 297.1335372 / [0.5 x 9.81] } = 7.783166671s
t_up + t_down = 16.22075113s....lol someone should check this, I'm not even doing this on a graphing calc...so my screen is limited..[/quote]
arent we looking for how long the rocket is in the air and not total time ?

I can't help but notice that basically every physics help thread on Basilmarket involves simple kinematics that ask you to work with specific parameters...

@nighttime: nooo...time up =/= time down
Okay time down is easier:
d=vi*t + 1/2 a*t^2
d= 297.133.....
a=+9.81m/s^2
vi=0
So we get t=sqrt{ 297.1335372 / [0.5 x 9.81] } = 7.783166671s
t_up + t_down = 16.22075113s....lol someone should check this, I'm not even doing this on a graphing calc...so my screen is limited..

[quote=Oyster]@nighttime: thinking about it again..."how long is it in the air" i think it means, we need to calculate the time falling back down to the ground.[/quote]

time it goes up and time it goes down should be the same same o-o

@nighttime: thinking about it again..."how long is it in the air" i think it means, we need to calculate the time falling back down to the ground.

Total distance = 297.1335372m

[quote=Oyster]Yeah, I thought at first, "how long is the rocket in the air" = "when does the rocket reach maximum height"
Isn't that the same....

Lol how many digits do you need, I'll redo it on the calcuator[/quote]
the max digits. SO is it the same?If so,then it is redundant o-o

"how long is the rocket in the air" we need to calculate the time, going back down to the ground.......

Lol how many digits do you need, I'll redo it on the calcuator

[quote=nighttime]yea i m down to like i 5 points or something[/quote]
is the distance the exact thing as shown in calculator ? and would the rocket staying in the air be the same as the time?
and oyster did got it right.
@Imgr8haha I was never good at physic.Only know the simple problems Physic is like my worst subject.

@nighttime: wow..maybe we should confirm this next time, before doing silly things..lol sorry i'm not actually writing this down, just doing all this on my calculator.

[quote=Oyster]i got t=8.437584455s and total height of 297.13m

btw, do you get points taken off for entering a wrong value every time?
I'm doing all this stuff on my calculator screen..maybe i should pull out a piece of paper..[/quote]

yea i m down to like i 5 points or something

i got t=8.437584455s and total height of 297.13m

btw, do you get points taken off for entering a wrong value every time?
I'm doing all this stuff on my calculator screen..maybe i should pull out a piece of paper..

[quote=ItzPootz]Oops! I forgot to divide by a in my spreadsheet, lol. t_total = 5.300847 s and max height is 248.8757 meters; unless I made another silly mistake.[/quote]

nope its wrong.
^tell me the answer oh wise one. Been at this for hrs. ._.

[quote=Oyster]Hmm.. well i used this formula to find vf: vf=sqrt{ vi^2 + 2*a*d} and I just plugged all the numbers in.
Then used -vf = -9.81*t

Then what is the exact number? Maybe you wrote the numbers wrong..
so what about 5.70sec?[/quote]
i typed the exact number 5.479591837. the site that i have to do my h.w. needs you to have exact number with no rounding.
@itzpootz wats the exact #? pleas tell me I really try to do it but fail x-x.

Hmm.. well i used this formula to find vf: vf=sqrt{ vi^2 + 2*a*d} and I just plugged all the numbers in.
Then used -vf = -9.81*t
OHHH that's right, you have to add the time together

[quote=Oyster]o.o I have 5.47s on my calc, my "vf=53.7m/s"[/quote]

i mess up somewhere then ._..5.47 isnt the answer either ._. Going give up. somehow manage to boost that 73% with other h.w.

o.o I have 5.47s on my calc, my "vf=53.7m/s"

[quote=Oyster]@nighttime: lol you'll get used to it. I always like to draw diagrams, it helps me. Well, I'm kinda word handicapped So I'm the visual type.[/quote]

i try to. Physics just drive me crazy. so it will take 5.700251726 sec to go to max?

@nighttime: lol you'll get used to it. I always like to draw diagrams, it helps me. Well, I'm kinda word handicapped So I'm the visual type.

Uhh...btw you [b]do[/b] know to use [b]-9.81[/b]m/s^2... for "a"
a is a vector, and the [b][-][/b] sign must be there, otherwise, there wouldn't be a "maximum height" it was just go off and off into space.

[quote=Oyster]vf2=0 (this is not directly stated, but [b]"maximum height"[/b] means vf=0) The object stops when it's reached it's maximum point and will go the opposite way.
So it makes sense vf2=0[/quote]
oh i forgot -facepalm- well use to seeing v0 and v for finding max height x-x

[quote=nighttime]but vf2 would be?[/quote]
vf2=0 (this is not directly stated, but [b]"maximum height"[/b] means vf=0) The object stops when it's reached it's maximum point and will go the opposite way.
So it makes sense vf2=0

i found t1 which is 2.960665278
ans vf=55.919
but vf2 would be?

[quote=nighttime]^ i use the 9.8 for this one? Now you want to solve for the time, t_2, it takes to reach maximum height ---> v_f2 = g*t_2 + v_f = 0[/quote]
Okay, so your ultimate goal is to calculate the maximum height your engine goes to.
There are two parts to it:
[1] your engine travels with a=2.07m/s^2 until s=150m
[2] then at 150m+ a=9.81m/s^2.

You want to calculate the "final" velocity of the engine while it is at s=150m. Which you can use this formula: vf^2 = vi^2 + 2*a*s
Vf = sqrt{vi^2 + 2*a*s} = sqrt{47.6^2 + 2*2.07*150} = 53.7m/s

Maybe you should draw a diagram, it helps you see where your engine is going and what you are trying to solve.

vf2 would be?

[quote=nighttime]do i use A= 9.8 at all?[/quote]
You use 9.81m/s^2 in the second part, after you found your vf when the engine reaches 150m.

Your question states:[quote=nighttime]It accelerates with a constant upward acceleration of 2.07 m/s^2 [b]until[/b] its engines stop at an altitude of 150m.[/quote]

do i use A= 9.8 at all?

@nighttime: s is disance
some calls it d, some calls it y..some s..

[quote=ItzPootz]I'll tell you the steps instead of working it all out for you:

1. Use the distance formula, s = 1/2a*t^2+v_i*t+s_i, and solve for the time, t, that it takes for the rocket to reach a height of 150m with a = 2.07 m/s^2 and an initial velocity of 47.6 m/s and an initial position s_i = 0. (It's a quadratic equation, use the positive value of t of course )

2. Use the time t you found to find the final velocity of the rocket when it reaches 150m; v_f = a*t + v_i

3. Now you want to solve for the time, t_2, it takes to reach maximum height ---> v_f2 = g*t_2 + v_f = 0

4. You use that time to find the maximum height with the distance formula s_max = 1/2g(t_2)^2 + v_f*(t_2) + s (you use the s from part 1 as the initial height)

Here's how I knew what to do: The rocket accelerates at a constant rate until it turns off at 150m. At that point it "coasts" upwards until gravity brings it to a stop (thus, v_f2 = 0). So, I needed to know how fast the rocket was going once it turned off at 150m. To find this, I needed to find out how much time it takes to get to that height, then I found the speed using the formula. After that point, you need to find the distance the rocket travels while slowing down due to gravity. To find this, you need to find out how long it took for the rocket to reach 0 velocity (the exact point before it starts falling down). So I use the velocity formula to find the time, then I use all the previously calculated values to solve for the final distance traveled.

Hope this helps [/quote]

what does s stand for? o-o