elp please!!!!

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Can someone please check question number 41 for me plz?!?!?!!?!?
can i suggest one thing

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it might look like its less work with the formulas and stuff, but there is a simpler way to think about these questions, even though the work might seem like its a bit more
i don't care, spit it out!!!!
okay look
i am.....^_^
|dw:1439951841864:dw|
ive already done that....how do i figureout the horizontal displacement of the projectile.....??? :)
now u know the horizontal speed Vx, so what we need to do is find the time it is in the air for, and multiply by that time
yes those are the initial vertical velocity and initial horizontal velocity.....
and to find how long its in Air for, we know the Vertival speed and we have the accerelation
one thing about the launch is that the starting and ending speed for the vertical is the same
jstu the direction is opposite
yup i know that.... :)
so what we need to do is see the time it takes for the velicity to go to 0 and mulitply by 2
our acc is 9.8 m/s/s so Vy= 35*sin 40 lets see how long that takes to go to 0 34*sin 40 - 9.8*t = 0 every second we are going to lose 9.8 see how many secs that will take find that time multiply by 2
35*sin 40 - 9.8*t = 0 t=35sin(40)/9.8 it will be up in air twice that time so our horizontal distanve must be ? Vx*2t = 35*cos(40) * 2*35sin(40)/9.8
comesout to 123.101
buddy that's not even closr to the real answer!!!
maybe some error somewhre xD
the answer is 40m.....
we're looking for the horizontal displacement of the ball....
so we figure out the time and plug that into the horizontal equation.... D = t/v or something like that....
here is the solution......wait
1 Attachment
actually this topic is slightly different, its about the chaning velocity and the angle....and to find the horzontal displacement we use the formula Dh = Vi2 X sin2O/a but i wanna know what is the Vi means here....
is it the initial horizontal velocity or the initial vertical velocity??????
dont use formulas just learn it
because i'm pluging in the number and both of my x and y comps are different :( cuz when i ply them into the equation i get a differentt answer :(
i think the answer there is wrong
it should be 123m ish
i wanna know what is the Vi2 means here????
vertical or horizontal......
what are u referring to again
hold up.....le me draw for you :)
wait i found the mistake they made
i think they plugged in wrong numbers or something because they ahve wrong horizontal and veritcal components
|dw:1439952832199:dw|
yes that is the right one
those are my numbers...when i plug them into their equation, i don't get the right answer????
did u get the method i was showing u up there?
i've tried both but i ain't getting 40m :"(
40 is wrong answer
yuss i already know that method.....
123 is right
you sure bra???
yes
i did it twice so it has to be right
well what do you get when you pplug in your findings into the formula, do you get 40?
let's try
i dont use formulas i dont know, id have to derive it
what formula do u want me to derive?
u want a formula with velocity and distance only?
:)
change in horzontal displacement = Vi^2 X Sin2 Theta/a
okay
|dw:1439953379986:dw|
normally i label my vertical values and horizontal values.... Vertical: Horizontal: a = -9.8m/sec2 dh = ? Dh = 0m/s Vh = 26.81m/sec Vi = 22.49m/s up t = ? Vf = 35m/sec t = ?
okay that is fine
that is good
now look maybe u can help work this out
wait why is there a Vf=35 that doesnt eexist
Vertical Vi=22.49m/s up Vf=22.49m/s down
now tell me how much time that will take with 9.8m/s^2 acc
i just need to find time, so for that i can use the following dy = Vi X t + 1/2(aXt^2) i'll get a qudratic expression and fro mthere i get my time....i ply that time into the horizontal equation = V = d/t => "t = d/V"
can u just answer this simpler qusetion first please
Vertical Vi=22.49m/s up Vf=22.49m/s down how much time will this take? from this Vi to Vf?
remember acceleration =V/t right
so what is the time?
a = V/t yes!! :D
so how much time will it take "Vertical Vi=22.49m/s up Vf=22.49m/s down how much time will this take? from this Vi to Vf? "
1 sec?
no
okay how much total change was there in the velocity
|dw:1439953983112:dw|
wait we have 2 velocities here Vi and Vf so which one am i going to use for that....??
|dw:1439953996826:dw|
accerelation = change in velocity / change in time
OMG i got it..... LOL
so how much did the time change by?
|dw:1439954092459:dw|
okay good
so this means it took 4.5 seconds for it go up and down
d= t/v
in this same time it was constantly moving horizontally at some speed
Horizontal speed * ~4.5 = distance in the horizontal v*t=d
pellet...-_-
its oka if the answer does not match the one in the book because that one is wrong, i can show u with a simple thought experiment
so 4.5s X 26.81m/s
yeah
121 LOL Yus!!! :"D
go online and look at a football being kicked in the air, see the approximate time
it just cannot be 40m meters that is too close
they will fire him if tahts the best he can kick it
this is an opening kick off
no but the thing is that, this chapter covers chaning angle and velocity...so to find the hori\ontal displacement .....they gave us the formula for that.....
i was just following the steps....
i see i see i get it
i wouldnt worry about all that stuff trust me, you should check out the MIt physics lectures they teach kinematics properly
There is only 2 real formulas u need to know for this whole course
d(t)= displacement as a function of time d'(t)= velocity, first derivative of displacement is velocity d''(t)=acceleration, 2nd der of displament is acc from these you can get all of kinematics equations
so what do i do in the exams cuz this chapter and the previous chapter literally had the same types of questions....where you have the final velocity, angle and maybe one or two thingsg from the horizontal.....but virtually exactly the same types of question.....
dont worry, for all those u just need some geometry and these principles
try some more questions, post it on here if u are stuck
cuz i picked a question from my previous chapter on "upward movign projectiles" and applied the formula from this chapter" Changing launch angle and initial velocity....the answer i got was totally different,......
ok thanks ....:")
stay cute!!! :)
wait u got 50 something right
yah.....
what u got there is the change in vertical distance
nononoono
sorry different question.....
i'll post it if i have the energy....
okk get to work!
eye eye captain....:"D

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