A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
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
 one year ago
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

This Question is Closed

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2your problem is a problem on a parabolic motion. Now your motion can be represented as below: dw:1439276629065:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now, 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
 one year ago
Best ResponseYou've already chosen the best response.0Okayyy will update you ma'am/sir if I have more question!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sooooooo I assume ycomponent will change but xcomponent will not because the acceleration goes only up to down

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2yes you are right that ycomponent of velocity will change but xcomponent of velocity will not change as there is no acceleration or force in the x direction

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2and 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 ydirection, once it reaches the highest point

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2so you can use the kinematical equation \[v = ut  (1/2) g t^2\] for the ycomponent 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 ydirection, once it reaches the highest point" the above equation i.e. \[v = ut  (1/2) g t^2\] is used to find 'velocity' in ydirection and you'll get it as a negative number but we want the positive one as we want the velocity in ydirection 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 ydirection or the ycomponent 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
 one year ago
Best ResponseYou've already chosen the best response.0Okay will read everything. I'll come back to you. Thankyou!

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2wel, thanks for the medal!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"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 ydirection, once it reaches the highest point" Is this theory? that If the ground is level, the speed up = speed down?

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2that can be proved and 'If the ground is level, the speed up = speed down' is same as that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0v=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

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2what is vo/ut are you dividing vo by ut??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm I'll use magnitude formula? soooo sqrt(7.056^2+14^2) = 15.68

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no I'm not dividing vo/t lol it means vo or u

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2yes then you are right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okayyyy but I'm still not convinced that you can use the kinematic equation to find the V

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you mean to say that v and ycomponent is the same?

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2and you can use the kinematical equation for the y component as the projectile is performing uniformly accelerated motion in the y direction;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okayyyy so moving on then. 7.056 is the ycomponent right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 14 is the xcomponent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0by which means I'll use arctan to get the angle

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2yes you are right but why do you want the angle??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\arctan(7.056/14)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought it is needed

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2i 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 ycomponent of the speed of the projectile it should be [v=ugt] where v is the speed in the y direction or the ycomponent of the speed

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2i am very very very very very sorry that i wasted your time on such a silly thing i am very sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okayyy let's start over lol soooo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yessss. I rereading the questionss

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so I'll use the formula v = ugt means u=0 g=9.8 t=1.2

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2okay now use the magnitude formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay sqrt(5.88^2+14^2)

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.0i did v= 01/2(9.8)(1.2) lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so angle will not be taken right?

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2yes angle will not be taken

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So moving on we got the magnitude

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll illustrate something is this correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.018.2838 is the magnitude right?

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2wait a second i'll calculate

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2yes thats the correct figure

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay okay ill illutrate something

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439310446738:dw

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2i'll show you the correct thing

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2well, you cannot show it in a graph like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay. Hmmm bc I got confused we already have a ycomponent = 14 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we need an xcomponent to get the magnitude

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2no we have the xcomponent=14 and y component before 1.2s = 11.76

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay wait let me clear this up I thought 14 is the ycomponent bc it is stated as maximum height

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. "

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2look 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
 one year ago
Best ResponseYou've already chosen the best response.2at 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you now we alraedy clear it up

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2okay! do you need help with the other parts of the problem??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we got the magnitude

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a.) is the magnitude right the 18.2838

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0same as b.) because of the theory

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that when the ground is level. they're equal to each other

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2yeah exactly you got it!!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0same as b.) but it will become negative bc it is accelearting downward?

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2it is accelerating downwards in both the cases

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u 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
 one year ago
Best ResponseYou've already chosen the best response.0and b.) is down bc it had reach the max height means it will freefall now

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2my 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
 one year ago
Best ResponseYou've already chosen the best response.0omgg im getting it now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm so dumb I realize it earlier but I forgot

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As 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
 one year ago
Best ResponseYou've already chosen the best response.0but when it moves downward the vertical velocity increases 9.8 m/s each second, until the object hits the ground.

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2yipeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee you got it!!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0god learning is fun hahahaha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0Heyyyy it is 1:08AM I'm a sluggish and sleepy we should continue tomorrow for more learning!! Thankyouuuuu

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2yeah and specially learning physics is funnnn!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but I think c,d,e,f will be a piece of cake right? Lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you and God bless!! Sorry if I'm hard to teached. I'm just homeschooling atm with the subject.

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2no you are not hard to teach this happens with everyone because this is a bit confusing! meet you tomorrow

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is the illustration right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So moving I'll try to understand C,D,E,F

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm attempting to finish the problem right now. Good luck to me lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0atm im attempting to get the max height

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Vy(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
 one year ago
Best ResponseYou've already chosen the best response.0to get the the t of max height 23.52m/s/9.8m/s^2 = 2.4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ill use the vot1/2gt = y to get the max height am I on the right track?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my final answer for (c,d,e,f) is (16.8,7.056,16.8,7.056)

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2First 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 "vot1/2gt = 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 ydirection 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
 one year ago
Best ResponseYou've already chosen the best response.0I got the correct answer @rajat97 woooooo everything is good. Thank you!!

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2oh! it's my pleasure to help you but did you understand whatever i said??

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2to 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
 one year ago
Best ResponseYou've already chosen the best response.0I take that as an advice sir. Will do!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm off finding more problemsss. Later!

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2please don't call me sir

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2and whenever you need help, please do remember me or there are lots of genius gems that openstudy has got!

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2okay then, bye i gotta go
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.