# Homework Help: 2d motion issues

1. Feb 9, 2010

### patterson

1. The problem statement, all variables and given/known data
I need to be able to get a equation to figure out the initial speed as an object leaves the origin, and I dont understand what exactly I am suppose to treat these as? Basically they are showing us vectors, Vo being velocity vector, and Vox and Voy being x and y components of that vector. If I am given a problem where I am given no Vox or Voy or Vo, and nothing else but what is given (acceleration is constant for y= -g and Ax=0) and distance from origin = 12.6m from apex of trajectory, how do I go about solving for Vox and Vot seeing that the book says I have to have Vo(Cos(Θo) = Vox and Vo(Sin(Θo) = Voy I am misunderstanding this because they say they are velocity, and i thought velocity is measured in m/s or ft/s as a speed, not a distance.

If I am only given a distance from the mid point (12.6m) and an apparent trajectory angle of 3 degrees, at what speed must the object leave its origin to rise .33m at its apex?

I am unsure what to do next because when I calculate the VoSinVo(Cos(Θo) = Vox and Vo(Sin(Θo) = Voy I have .66m for Voy and 12.617 for Vo and this is obviously not right
How the heck am I suppose to pull speed from this anyways I just dont get this crap ive worked 10 hours on this and I just dont get it

2. Relevant equations
Vo(Cos(Θo) = Vox and Vo(Sin(Θo) = Voy
Vx=Voxt + Ax(squared)
∆x = Voxt + 1/2 axt(squared)
Vx(squared) = Voxt(squared) + 2ax∆x
Vy=Voy + Ayt
∆y = Voyt + 1/2 Ayt(squared)
Vy(squared) = Voy2 + 2ay∆y

3. The attempt at a solution

where do I begin to explain what I am doing here I have Vox= VoCos(theta) and Voy=VoSin(theta)
12.6m Sin (3degrees) = .66 = Voyt
Asquared + bsquared = Csquared
.66(squared) + 12.6m(squared) = C = 12.617m/s
so Vo= 12.617m/s
∆x = Voxt = 25.2m = 12.6m(t)
t= 2 seconds
∆y = y=yo = Voyt - 1/2gt(squared)
.66m = .66m/s (t) - (4.9m/s(squared)t(squared)
Vx = Vox = VoCos(theta) = 12.6m/s
Vy = VoSin (theta) - gt
12.617 sin (3degrees) - (9.8m/s(squared))(2seconds) = - 18.94 m/s

see what I mean this book sucks the teacher helps out nothing but more confusion, no clarification. I am going to blow up soon and i just need more understanding how this stuff is even suppose to flow together, they give me no formulas to work with I try to use their formulas and look what I come up with nothing but stuff that is wrong all the way through apparently according to the book which tlels me it is 48.6 m/s
how the heck do they come up with that I have -18.94m/s
i mean seriosuly if they are going to put questions like this in here why do they not even tell us how to even go about getting a velocity from nothing but x and y and a degree i mean speed from that i dont get it someone please help before I off myself.

2. Feb 9, 2010

### rl.bhat

Post the problem in exact words. What is given and what is required.

3. Feb 9, 2010

### thebigstar25

ooooooooh dont confuse yourself this way, dont worry me and the other members will help you out through your problem..

See, mistake number(1) you used Vo=12.6 which is obviously wrong assumption,why? Because this 12.6 represents a distance , you have said it yourself Vo is velocity and should be measured as m/s not m ..

Im not sure if i have got the complete vision of your question :S .. Can you please write the exact problem as written in your book so I can help you better :) ..

4. Feb 9, 2010

### patterson

ok I will write one of many problems which has me confused as heck.
teacher is tryin to show me how to solve them but i have no idea what he is even doing, he is drawing 6 - 9 different x and y graphs for Vy,Vx, X, Y, and Ax, Ay and from there he does not even know anything about the problem he just starts pulling points off the graphs and starts calling them like (t, V(t) and telling me to throw other equations into that and solving for that but I have absolute no idea how he gets it he claims that acceleration area graph so whatever is inside the box is the position which makes no sense to me, then he starts doing different stuff anyway I am lost I try and try and try and more then anything i hate the fact that I cant learn this stuff because I dont get it and I try to get it....

5. Feb 9, 2010

### patterson

A tennis player standing 12.6 m from the net hits the ball at 3° above the horizontal. To clear the net the ball must rise at least .330m. If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racket?

ok so 12.6m = x and not Vox, and .33m = y, and not voy. Given the 3 degrees and that is it.

now the teacher dont let us use formulas on this stuff, so basically I have to draw these graphs, label points like Vox Voy Vo and then solve for Vo or V i dont evne know the difference between them, and solving for other things like the time. Accel is 0 in the Ax coordinate system, and the Ay Coordinate system is -9.8m/s squared. Velocity in the constand direction for both x and y. I just dont get how I am suppose to draw these graphs, pull coordinates off them like Vsub T and t,V(t) he calls it T, V(of t) and apparently these are functions of something I dont understand anyway

Hopefully you guys can show me some insight on how the heck I am suppose to go about getting the 2 different 6 equations for the two unknowns or whatever i just dont get it, if i am not given vo, vox, or voy, how do i even get any of those without having any information ubt what i know and I dont know anything about this crap seriously upsets me, i mean come on, i can program in every fricken language out there, and write asp.NET and im just too dang stupid to understand physics, living in a virtual world where I declare constants and variables and changes in manipulations using statements of selection, and conditions, along with nested statements.... just hate this physics stuff because I like to get a thourough understanding of stuff before i just try and figure it out doing things like I already have been doing for all the problems ive solved... homework is due tomorrow and the exam is due tomorrow I staedup til 4 am last nite trying to figure it out and of course no go.. so im unsure whether to just jump off the earth and say screw my degree in computer engineering.

6. Feb 9, 2010

### rl.bhat

Let Vo be the velocity of projection of the ball which makes an angle θ degrees with x-axis.
The ball moves horizontally and vertically simultaneously.
Resolve Vo into two components.
Horizontal component = Vo*cosθ
Vertical component = Vo*sinθ.
Since there is no acceleration along x-axis, Vo*cosθ remains constant.
In the vertical direction, Vo*sinθ decreases and becomes zero at the apex.
So in the vertical direction, initial velocity is gives, final velocity is given, acceleration is Known( -9.8 m/s^2) and displacement is given.( 0.330m).
In the horizontal direction, velocity is Vo*cosθ and displacement is given. (12.6 m).
From this information you can find the time t taken by the ball to reach the net.( t = d/Vo*cosθ).
For vertical displacement, use appropriate kinematic equation and substitute the value of t=d/Vo*cosθ. Simplify and solve the quadratic to find Vo.

7. Feb 10, 2010

### thebigstar25

I think that the problem looks something like this:

http://img6.imageshack.us/img6/8743/64523207.jpg [Broken]

you are given x=12.6 m , y= 0.33m , and theta = 3 degree , and you are asked to find the initial velocity vo , so what do you have to do?

you know that vo = sqrt( vox^2 + voy^2), so if you find vox and voy you can find out vo ..

as rl.bhat said , vox = vocostheta and voy=vosintheta , you have theta but still you can see that is not enough ..

so what you have to do next? , at this point and before you give up return to the question and see if you can somehow extract any hidden information..

you know that this ball will cover 12.6 m in the x-axis , so this 12.6 represents the max. distance(range) and there is a useful equation you can use here :

the range(max. x) = vo*costheta*t(total) ,(which is simply distance= velocity*time)

substituting in this equation you will know the value of vo*t(total)

so what to do here? , you may realize that if you have one more equation you can then get your work done ..

lets now move to the y-axis , what do we know ? you said that the acceleration is constant , so you have many equations to use:

(vfy) = (voy) -g*t
(vfy)^2 = (voy)^2 - 2gy
y = (voy)t -0.5gt^2 , where vfy = final velocity in y-axis , voy=initial velocity in x-axis

is it clear enough? can you go from there? if it is still not clear ,please ask again..

Last edited by a moderator: May 4, 2017
8. Feb 11, 2010

### patterson

My problem is that we have to draw out 5 - 6 graphs for each of these lines for position vs time, accel vs time, and velocity vs time, of both Y and X directions.

Then they are just ranomly calling tings points like T,V(T) or V(V(x)) i dont get how the heck they are going and just grabbing points off the graph, then by using these points they say forinstance call the Y-accel graph (-9.8) then somehow they have that as a function of something i dont get and then they have -9.8t then they plug this into another F(x)) that i dont get any of it, and this kid in my class is telling me that the formulas dont matter that i am not going to get credit unless I pull the stuff off of the graphs and then substitute those back into each other to solve for w/e but if that is the case why is it that he makes it so confusing to me so velocity curve is a right triangle, 1/2base times height to get the area of the velocity which is equal to the change in position? I have no idea, all I know is that I am seriously sick of this crap already

9. Feb 11, 2010

### thebigstar25

now im the one who got confused >_<" .. are you supposed to do the graphs using a certian language? what is this course?