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

- anonymous

- schrodinger

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- Michele_Laino

your problem is a problem on a parabolic motion.
Now your motion can be represented as below:
|dw:1439276629065:dw|

- 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

Okayyy will update you ma'am/sir if I have more question!

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## More answers

- anonymous

I got confuse

- anonymous

Sooooooo I assume y-component will change but x-component will not because the acceleration goes only up to down

- 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

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

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

Okay will read everything. I'll come back to you. Thankyou!

- rajat97

wel, thanks for the medal!

- 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

that can be proved and 'If the ground is level, the speed up = speed down' is same as that

- anonymous

v=utâˆ’(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

I got -7.056

- rajat97

what is vo/ut are you dividing vo by ut??

- anonymous

Hmmm I'll use magnitude formula? soooo sqrt(7.056^2+14^2) = 15.68

- rajat97

yes

- anonymous

no I'm not dividing vo/t lol it means vo or u

- rajat97

yes then you are right

- anonymous

Okayyyy but I'm still not convinced that you can use the kinematic equation to find the V

- anonymous

OH WAIT

- anonymous

you mean to say that v and y-component is the same?

- rajat97

wohooo! you got it

- 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

okayyyy so moving on then. 7.056 is the y-component right?

- anonymous

and 14 is the x-component

- anonymous

by which means I'll use arctan to get the angle

- rajat97

yes you are right but why do you want the angle??

- anonymous

\[\arctan(-7.056/14)\]

- anonymous

Oh lol

- anonymous

I thought it is needed

- 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

i am very very very very very sorry that i wasted your time on such a silly thing i am very sorry

- anonymous

Okayyy let's start over lol soooo

- anonymous

I'll use 1.2s = t

- rajat97

yes

- rajat97

you there?

- anonymous

Yessss. I rereading the questionss

- rajat97

okay

- anonymous

okay so I'll use the formula v = u-gt means u=0 g=9.8 t=1.2

- anonymous

5.88

- rajat97

okay now use the magnitude formula

- anonymous

okay sqrt(5.88^2+14^2)

- anonymous

15.1847

- 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

omg okay sorry.

- anonymous

i did v= 0-1/2(9.8)(1.2) lol

- rajat97

no matter man :)

- anonymous

So okay moving on

- anonymous

so angle will not be taken right?

- rajat97

yes angle will not be taken

- anonymous

So moving on we got the magnitude

- rajat97

yes we get it now

- anonymous

I'll illustrate something is this correct

- rajat97

okay!

- anonymous

18.2838 is the magnitude right?

- rajat97

wait a second i'll calculate

- anonymous

sqrt(11.76^2+14^2)

- rajat97

yes thats the correct figure

- anonymous

okay okay ill illutrate something

- anonymous

|dw:1439310446738:dw|

- rajat97

i'll show you the correct thing

- anonymous

okaaaaay thank you

- rajat97

well, you cannot show it in a graph like this

- anonymous

okay. Hmmm bc I got confused we already have a y-component = 14 right?

- anonymous

so we need an x-component to get the magnitude

- rajat97

no we have the x-component=14 and y component before 1.2s = 11.76

- anonymous

Oh okay wait let me clear this up I thought 14 is the y-component bc it is stated as maximum height

- anonymous

oh wait hmm

- anonymous

i think im wrong

- anonymous

"The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. "

- 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

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

OKAY GOT IT.

- anonymous

Thank you now we alraedy clear it up

- rajat97

okay!
do you need help with the other parts of the problem??

- anonymous

So we got the magnitude

- anonymous

I think yes?

- anonymous

a.) is the magnitude right the 18.2838

- anonymous

same as b.) because of the theory

- anonymous

that when the ground is level. they're equal to each other

- rajat97

yeah exactly you got it!!!!

- anonymous

same as b.) but it will become negative bc it is accelearting downward?

- anonymous

or nah

- rajat97

it is accelerating downwards in both the cases

- 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

and b.) is down bc it had reach the max height means it will freefall now

- 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

omgg im getting it now

- anonymous

I'm so dumb I realize it earlier but I forgot

- 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

but when it moves downward the vertical velocity increases 9.8 m/s each second, until the object hits the ground.

- rajat97

yipeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee you got it!!!!

- anonymous

god learning is fun hahahaha

- 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

Heyyyy it is 1:08AM I'm a sluggish and sleepy we should continue tomorrow for more learning!! Thankyouuuuu

- rajat97

yeah and specially learning physics is funnnn!

- anonymous

but I think c,d,e,f will be a piece of cake right? Lol

- rajat97

no problem !

- anonymous

Thank you and God bless!! Sorry if I'm hard to teached. I'm just homeschooling atm with the subject.

- rajat97

no you are not hard to teach
this happens with everyone because this is a bit confusing!
meet you tomorrow

- anonymous

##### 1 Attachment

- anonymous

This is the illustration right?

- anonymous

So moving I'll try to understand C,D,E,F

- anonymous

moving on*

- anonymous

I'm attempting to finish the problem right now. Good luck to me lol

- anonymous

atm im attempting to get the max height

- 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

Vy(1.2)*

- anonymous

to get the the t of max height 23.52m/s/9.8m/s^2 = 2.4

- anonymous

t=v/a

- anonymous

ill use the vot-1/2gt = y to get the max height am I on the right track?

- anonymous

which is 28.224

- anonymous

my final answer for (c,d,e,f) is (-16.8,-7.056,16.8,-7.056)

- anonymous

idk if I'm right

- 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

All right then.

- anonymous

I got the correct answer @rajat97 woooooo everything is good. Thank you!!

- rajat97

oh! it's my pleasure to help you
but did you understand whatever i said??

- anonymous

Yess!!

- 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

I take that as an advice sir. Will do!!

- anonymous

I'm off finding more problemsss. Later!

- rajat97

please don't call me sir

- rajat97

and whenever you need help, please do remember me or there are lots of genius gems that openstudy has got!

- rajat97

okay then, bye
i gotta go

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