anonymous
  • anonymous
The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. (a) What is the magnitude of the velocity of the projectile 1.2 s before it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.2 s after it achieves its maximum height? If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile 1.2 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate
Physics
schrodinger
  • schrodinger
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Michele_Laino
  • Michele_Laino
your problem is a problem on a parabolic motion. Now your motion can be represented as below: |dw:1439276629065:dw|
Michele_Laino
  • Michele_Laino
now, the horizontal component of the velocity is constant, and in general we can write this: \[\Large \begin{gathered} {v_x}\left( t \right) = {V_0}\cos \theta \hfill \\ \hfill \\ {v_y}\left( t \right) = {V_0}\sin \theta - gt \hfill \\ \end{gathered} \] where v_x and v_y are the components of the velocity at a generic time t, and g is the gravity
anonymous
  • anonymous
Okayyy will update you ma'am/sir if I have more question!

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anonymous
  • anonymous
I got confuse
anonymous
  • anonymous
Sooooooo I assume y-component will change but x-component will not because the acceleration goes only up to down
rajat97
  • rajat97
yes you are right that y-component of velocity will change but x-component of velocity will not change as there is no acceleration or force in the x direction
rajat97
  • rajat97
and the speed with wich the projectile will descend after time 't' is equal to the speed with wich the projectile was ascending before time 't' in the y-direction, once it reaches the highest point
rajat97
  • rajat97
so you can use the kinematical equation \[v = ut - (1/2) g t^2\] for the y-component of velocity where u=0 and we know g=10m/s^2 or 9.8m/s^2 (whatever is specified in the question or thats just a matter of our assumption) and t=1.2 from this equation you will get the speed with which the projectile will descend in the y direction after time 1.2s and reffering to the previous post, as i said, that "The speed with wich the projectile will descend after time 't' is equal to the speed with wich the projectile was ascending before time 't' in the y-direction, once it reaches the highest point" the above equation i.e. \[v = ut - (1/2) g t^2\] is used to find 'velocity' in y-direction and you'll get it as a negative number but we want the positive one as we want the velocity in y-direction when the projectile is ascending which we consider to be positive due to our sign convention now, once you get 'v' i.e. the speed on the y-direction or the y-component of the velocity before 1.2 s, you can find the magnitude of velocity by using the formula for magnitude of resultant of two vectors at right angles to each other (if a abd b are two vectors, perpendicular to each other, the magnitude of their resultant will be sqrt.[(a^2)+(b^2)]) solve this part and then i'll help you with the other parts of the question
anonymous
  • anonymous
Okay will read everything. I'll come back to you. Thankyou!
rajat97
  • rajat97
wel, thanks for the medal!
anonymous
  • anonymous
"and the speed with wich the projectile will descend after time 't' is equal to the speed with wich the projectile was ascending before time 't' in the y-direction, once it reaches the highest point" Is this theory? that If the ground is level, the speed up = speed down?
rajat97
  • rajat97
that can be proved and 'If the ground is level, the speed up = speed down' is same as that
anonymous
  • anonymous
v=ut−(1/2)gt^2 is the formula you stated right? because I thought the formula is equate to y=vo/u(t)-(1/2)gt^2
anonymous
  • anonymous
I got -7.056
rajat97
  • rajat97
what is vo/ut are you dividing vo by ut??
anonymous
  • anonymous
Hmmm I'll use magnitude formula? soooo sqrt(7.056^2+14^2) = 15.68
rajat97
  • rajat97
yes
anonymous
  • anonymous
no I'm not dividing vo/t lol it means vo or u
rajat97
  • rajat97
yes then you are right
anonymous
  • anonymous
Okayyyy but I'm still not convinced that you can use the kinematic equation to find the V
anonymous
  • anonymous
OH WAIT
anonymous
  • anonymous
you mean to say that v and y-component is the same?
rajat97
  • rajat97
wohooo! you got it
rajat97
  • rajat97
and you can use the kinematical equation for the y component as the projectile is performing uniformly accelerated motion in the y direction;)
anonymous
  • anonymous
okayyyy so moving on then. 7.056 is the y-component right?
anonymous
  • anonymous
and 14 is the x-component
anonymous
  • anonymous
by which means I'll use arctan to get the angle
rajat97
  • rajat97
yes you are right but why do you want the angle??
anonymous
  • anonymous
\[\arctan(-7.056/14)\]
anonymous
  • anonymous
Oh lol
anonymous
  • anonymous
I thought it is needed
rajat97
  • rajat97
i am the dumbest person in the world i don't know what i was thinking about but i've posted the wron formula for the y-component of the speed of the projectile it should be [v=u-gt] where v is the speed in the y direction or the y-component of the speed
rajat97
  • rajat97
i am very very very very very sorry that i wasted your time on such a silly thing i am very sorry
anonymous
  • anonymous
Okayyy let's start over lol soooo
anonymous
  • anonymous
I'll use 1.2s = t
rajat97
  • rajat97
yes
rajat97
  • rajat97
you there?
anonymous
  • anonymous
Yessss. I rereading the questionss
rajat97
  • rajat97
okay
anonymous
  • anonymous
okay so I'll use the formula v = u-gt means u=0 g=9.8 t=1.2
anonymous
  • anonymous
5.88
rajat97
  • rajat97
okay now use the magnitude formula
anonymous
  • anonymous
okay sqrt(5.88^2+14^2)
anonymous
  • anonymous
15.1847
rajat97
  • rajat97
i think you have done something wrong with the y component of velocity it should be v = 0 - 9.8 x 1.2 = -11.76m/s (we want the positive one so 11.76m/s)
anonymous
  • anonymous
omg okay sorry.
anonymous
  • anonymous
i did v= 0-1/2(9.8)(1.2) lol
rajat97
  • rajat97
no matter man :)
anonymous
  • anonymous
So okay moving on
anonymous
  • anonymous
so angle will not be taken right?
rajat97
  • rajat97
yes angle will not be taken
anonymous
  • anonymous
So moving on we got the magnitude
rajat97
  • rajat97
yes we get it now
anonymous
  • anonymous
I'll illustrate something is this correct
rajat97
  • rajat97
okay!
anonymous
  • anonymous
18.2838 is the magnitude right?
rajat97
  • rajat97
wait a second i'll calculate
anonymous
  • anonymous
sqrt(11.76^2+14^2)
rajat97
  • rajat97
yes thats the correct figure
anonymous
  • anonymous
okay okay ill illutrate something
anonymous
  • anonymous
|dw:1439310446738:dw|
rajat97
  • rajat97
i'll show you the correct thing
anonymous
  • anonymous
okaaaaay thank you
rajat97
  • rajat97
well, you cannot show it in a graph like this
anonymous
  • anonymous
okay. Hmmm bc I got confused we already have a y-component = 14 right?
anonymous
  • anonymous
so we need an x-component to get the magnitude
rajat97
  • rajat97
no we have the x-component=14 and y component before 1.2s = 11.76
anonymous
  • anonymous
Oh okay wait let me clear this up I thought 14 is the y-component bc it is stated as maximum height
anonymous
  • anonymous
oh wait hmm
anonymous
  • anonymous
i think im wrong
anonymous
  • anonymous
"The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. "
rajat97
  • rajat97
look i'll explain you what exactly happens in projectile motion: First of all we need to clear that during the projectile motion, the only force acting on the object is the gravitational force and that too in the vertically downwards direction there is no force in the horizontal direction and we call the vertical direction as the y direction and the horizontal direction as the x direction as there is no force acting on the object in the x direction, the speed of the object will not change in the x direction (Which is 14 here) and the object is performing uniformly accelerated motion in the y direction (you can call it motion under gravity) so we use the kinematical equations for uniformly accelerated motion if we want to find any of the parameters related with the y direction and we use the formula displacement=velocity x time and it's other forms to find any parameters related to the horizontal i.e. the x direction
rajat97
  • rajat97
at maximum height, just as motion under gravity, the speed of the object in the y direction is 0 so the speed contributing to the total speed of the object at the heighest point of the projectile motion is the speed in the x direction and here it is 14
anonymous
  • anonymous
OKAY GOT IT.
anonymous
  • anonymous
Thank you now we alraedy clear it up
rajat97
  • rajat97
okay! do you need help with the other parts of the problem??
anonymous
  • anonymous
So we got the magnitude
anonymous
  • anonymous
I think yes?
anonymous
  • anonymous
a.) is the magnitude right the 18.2838
anonymous
  • anonymous
same as b.) because of the theory
anonymous
  • anonymous
that when the ground is level. they're equal to each other
rajat97
  • rajat97
yeah exactly you got it!!!!
anonymous
  • anonymous
same as b.) but it will become negative bc it is accelearting downward?
anonymous
  • anonymous
or nah
rajat97
  • rajat97
it is accelerating downwards in both the cases
anonymous
  • anonymous
u sure I thought a.) means upward because it is stated it is before reaching max height, which i thought it is still going up to reach its destination
anonymous
  • anonymous
and b.) is down bc it had reach the max height means it will freefall now
rajat97
  • rajat97
my internet connection has some problems so i may seem to be offline but i'm not look, the force acting on the object is the force of gravitation which always acts in the downward direction. we know that the acceleration is always directed in the direction of the force thus the acceleration is always downwards if the acceleration would upwards, the projectile would fly up, up and up and never come back but the speed and the acceleration of the object are directed in opposite directions when the object is rising. thus the speed of the object decreases and finally becomes '0' at the MAXIMUM HEIGHT
anonymous
  • anonymous
omgg im getting it now
anonymous
  • anonymous
I'm so dumb I realize it earlier but I forgot
anonymous
  • anonymous
As an object moves upward, the vertical component of its velocity decreases 9.8 m/s each second. At the maximum height, the vertical velocity = 0 m/s
anonymous
  • anonymous
but when it moves downward the vertical velocity increases 9.8 m/s each second, until the object hits the ground.
rajat97
  • rajat97
yipeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee you got it!!!!
anonymous
  • anonymous
god learning is fun hahahaha
anonymous
  • anonymous
so what's next is 18.2838 the answer for a and b ? if so. we move to c,d,e,f if you want.
anonymous
  • anonymous
Heyyyy it is 1:08AM I'm a sluggish and sleepy we should continue tomorrow for more learning!! Thankyouuuuu
rajat97
  • rajat97
yeah and specially learning physics is funnnn!
anonymous
  • anonymous
but I think c,d,e,f will be a piece of cake right? Lol
rajat97
  • rajat97
no problem !
anonymous
  • anonymous
Thank you and God bless!! Sorry if I'm hard to teached. I'm just homeschooling atm with the subject.
rajat97
  • rajat97
no you are not hard to teach this happens with everyone because this is a bit confusing! meet you tomorrow
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
This is the illustration right?
anonymous
  • anonymous
So moving I'll try to understand C,D,E,F
anonymous
  • anonymous
moving on*
anonymous
  • anonymous
I'm attempting to finish the problem right now. Good luck to me lol
anonymous
  • anonymous
atm im attempting to get the max height
anonymous
  • anonymous
Vy(1.4) = 11.76m/s = Vyi - g*1.2 = Vyi - 9.8*1.2 => Vyi = 11.76+11.76 = 23.52m/s up at launch.
anonymous
  • anonymous
Vy(1.2)*
anonymous
  • anonymous
to get the the t of max height 23.52m/s/9.8m/s^2 = 2.4
anonymous
  • anonymous
t=v/a
anonymous
  • anonymous
ill use the vot-1/2gt = y to get the max height am I on the right track?
anonymous
  • anonymous
which is 28.224
anonymous
  • anonymous
my final answer for (c,d,e,f) is (-16.8,-7.056,16.8,-7.056)
anonymous
  • anonymous
idk if I'm right
rajat97
  • rajat97
First of all, thanks for the medal @peachpi well, @arvnoodle , the graph that you have uploaded is correct for the position of the particle but not for it's velocity. Next, you want to use the formula y = vo(t) - (1/2)gt^2 and not "vot-1/2gt = y" and we do not need to find the maximum height (The particle or the object is currently at the maximum height or the topmost point of the projectile) by using the formula y = vo(t) - (1/2)gt^2 , you'll get the position of the particle 1.2 s before it was at the max. height, if you put vo(initial component of velocity in the y-direction which is 0 at the topmost point) = 0, g = 9.8 and t = -1.2(time is negative as we have to find the past situation) you'll get a negative value, which suggests that the particle was below the position to that it is now on and for the x direction, you should use the formula x=vt and put v = 14 and t = -1.2
anonymous
  • anonymous
All right then.
anonymous
  • anonymous
I got the correct answer @rajat97 woooooo everything is good. Thank you!!
rajat97
  • rajat97
oh! it's my pleasure to help you but did you understand whatever i said??
anonymous
  • anonymous
Yess!!
rajat97
  • rajat97
to solve projectile motion problems, always break the motion into x and y directions and then solve for them individually this may be lenghty but way too better than those projectile motion formulae!
anonymous
  • anonymous
I take that as an advice sir. Will do!!
anonymous
  • anonymous
I'm off finding more problemsss. Later!
rajat97
  • rajat97
please don't call me sir
rajat97
  • rajat97
and whenever you need help, please do remember me or there are lots of genius gems that openstudy has got!
rajat97
  • rajat97
okay then, bye i gotta go

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