YumYum247
  • YumYum247
elp please!!!!
Physics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
YumYum247
  • YumYum247
Can someone please check question number 41 for me plz?!?!?!!?!?
YumYum247
  • YumYum247
@dan815
dan815
  • dan815
can i suggest one thing

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More answers

dan815
  • dan815
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
YumYum247
  • YumYum247
i don't care, spit it out!!!!
dan815
  • dan815
okay look
YumYum247
  • YumYum247
i am.....^_^
dan815
  • dan815
|dw:1439951841864:dw|
YumYum247
  • YumYum247
ive already done that....how do i figureout the horizontal displacement of the projectile.....??? :)
dan815
  • dan815
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
YumYum247
  • YumYum247
yes those are the initial vertical velocity and initial horizontal velocity.....
dan815
  • dan815
and to find how long its in Air for, we know the Vertival speed and we have the accerelation
dan815
  • dan815
one thing about the launch is that the starting and ending speed for the vertical is the same
dan815
  • dan815
jstu the direction is opposite
YumYum247
  • YumYum247
yup i know that.... :)
dan815
  • dan815
so what we need to do is see the time it takes for the velicity to go to 0 and mulitply by 2
dan815
  • dan815
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
dan815
  • dan815
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
dan815
  • dan815
comesout to 123.101
YumYum247
  • YumYum247
buddy that's not even closr to the real answer!!!
dan815
  • dan815
maybe some error somewhre xD
YumYum247
  • YumYum247
the answer is 40m.....
YumYum247
  • YumYum247
we're looking for the horizontal displacement of the ball....
YumYum247
  • YumYum247
so we figure out the time and plug that into the horizontal equation.... D = t/v or something like that....
YumYum247
  • YumYum247
here is the solution......wait
YumYum247
  • YumYum247
1 Attachment
YumYum247
  • YumYum247
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....
YumYum247
  • YumYum247
is it the initial horizontal velocity or the initial vertical velocity??????
dan815
  • dan815
dont use formulas just learn it
YumYum247
  • YumYum247
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 :(
dan815
  • dan815
i think the answer there is wrong
dan815
  • dan815
it should be 123m ish
YumYum247
  • YumYum247
i wanna know what is the Vi2 means here????
YumYum247
  • YumYum247
vertical or horizontal......
dan815
  • dan815
what are u referring to again
YumYum247
  • YumYum247
hold up.....le me draw for you :)
dan815
  • dan815
wait i found the mistake they made
dan815
  • dan815
i think they plugged in wrong numbers or something because they ahve wrong horizontal and veritcal components
YumYum247
  • YumYum247
|dw:1439952832199:dw|
dan815
  • dan815
yes that is the right one
YumYum247
  • YumYum247
those are my numbers...when i plug them into their equation, i don't get the right answer????
dan815
  • dan815
did u get the method i was showing u up there?
YumYum247
  • YumYum247
i've tried both but i ain't getting 40m :"(
dan815
  • dan815
40 is wrong answer
YumYum247
  • YumYum247
yuss i already know that method.....
dan815
  • dan815
123 is right
YumYum247
  • YumYum247
you sure bra???
dan815
  • dan815
yes
dan815
  • dan815
i did it twice so it has to be right
YumYum247
  • YumYum247
well what do you get when you pplug in your findings into the formula, do you get 40?
YumYum247
  • YumYum247
let's try
dan815
  • dan815
i dont use formulas i dont know, id have to derive it
dan815
  • dan815
what formula do u want me to derive?
dan815
  • dan815
u want a formula with velocity and distance only?
dan815
  • dan815
:)
YumYum247
  • YumYum247
change in horzontal displacement = Vi^2 X Sin2 Theta/a
dan815
  • dan815
okay
dan815
  • dan815
|dw:1439953379986:dw|
YumYum247
  • YumYum247
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 = ?
dan815
  • dan815
okay that is fine
dan815
  • dan815
that is good
dan815
  • dan815
now look maybe u can help work this out
dan815
  • dan815
wait why is there a Vf=35 that doesnt eexist
dan815
  • dan815
Vertical Vi=22.49m/s up Vf=22.49m/s down
dan815
  • dan815
now tell me how much time that will take with 9.8m/s^2 acc
YumYum247
  • YumYum247
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"
dan815
  • dan815
can u just answer this simpler qusetion first please
dan815
  • dan815
Vertical Vi=22.49m/s up Vf=22.49m/s down how much time will this take? from this Vi to Vf?
dan815
  • dan815
remember acceleration =V/t right
dan815
  • dan815
so what is the time?
YumYum247
  • YumYum247
a = V/t yes!! :D
dan815
  • dan815
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? "
YumYum247
  • YumYum247
1 sec?
dan815
  • dan815
no
dan815
  • dan815
okay how much total change was there in the velocity
dan815
  • dan815
|dw:1439953983112:dw|
YumYum247
  • YumYum247
wait we have 2 velocities here Vi and Vf so which one am i going to use for that....??
dan815
  • dan815
|dw:1439953996826:dw|
dan815
  • dan815
accerelation = change in velocity / change in time
YumYum247
  • YumYum247
OMG i got it..... LOL
dan815
  • dan815
so how much did the time change by?
YumYum247
  • YumYum247
|dw:1439954092459:dw|
dan815
  • dan815
okay good
dan815
  • dan815
so this means it took 4.5 seconds for it go up and down
YumYum247
  • YumYum247
d= t/v
dan815
  • dan815
in this same time it was constantly moving horizontally at some speed
dan815
  • dan815
Horizontal speed * ~4.5 = distance in the horizontal v*t=d
YumYum247
  • YumYum247
pellet...-_-
dan815
  • dan815
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
YumYum247
  • YumYum247
so 4.5s X 26.81m/s
dan815
  • dan815
yeah
YumYum247
  • YumYum247
121 LOL Yus!!! :"D
dan815
  • dan815
go online and look at a football being kicked in the air, see the approximate time
dan815
  • dan815
it just cannot be 40m meters that is too close
dan815
  • dan815
they will fire him if tahts the best he can kick it
dan815
  • dan815
this is an opening kick off
YumYum247
  • YumYum247
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.....
YumYum247
  • YumYum247
i was just following the steps....
dan815
  • dan815
i see i see i get it
dan815
  • dan815
i wouldnt worry about all that stuff trust me, you should check out the MIt physics lectures they teach kinematics properly
dan815
  • dan815
There is only 2 real formulas u need to know for this whole course
dan815
  • dan815
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
YumYum247
  • YumYum247
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.....
dan815
  • dan815
dont worry, for all those u just need some geometry and these principles
dan815
  • dan815
try some more questions, post it on here if u are stuck
YumYum247
  • YumYum247
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,......
YumYum247
  • YumYum247
ok thanks ....:")
YumYum247
  • YumYum247
stay cute!!! :)
dan815
  • dan815
wait u got 50 something right
YumYum247
  • YumYum247
yah.....
dan815
  • dan815
what u got there is the change in vertical distance
YumYum247
  • YumYum247
nononoono
YumYum247
  • YumYum247
sorry different question.....
YumYum247
  • YumYum247
i'll post it if i have the energy....
dan815
  • dan815
okk get to work!
YumYum247
  • YumYum247
eye eye captain....:"D

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