Ball trajectory velocity and displacement vectors

[late347]

Homework Statement

ball is thrown at slighta angle upwards and to the right. Angle between horizontal and the launch velocity vector is 35deg upwards.
Launch velocity vector (v_0) has magnitude of 12m/s
Only gravity affects the ball as a force. (no air resistance)
t_0 = 0 secs
t_1 = 0.2 secs

1. calculate what the velocity ($v_1$) is 0,2s after the launch.
2. work from the previous scenario and what is the displacement of the ball, ostensibly between the time period of $t_0=0s$ and $t_1=0,2s$

Homework Equations

I tryt to use the relevant equations as I go along.

• vector components
• v_1=v_0+at
• trigonometric functions

The Attempt at a Solution

I was quite a bit confused conceptually speaking about the total displacement between the time period of 0.2s from the 0s timepoint onward.1. velocity

according to my notes, it should be so that the $\vec{v_1}$ can be constructed from components.
it was known in the beginning that $V_{0x}= V_{1x}$ because the only force affecting is in the vertical direction from gravity. Thus I think the x direction is unaffected by a force. y direction does have gravity though.

In the y direction, we have originally $V_{0y} = sin(35)*|\vec{V_0}|~~=6.8829m/s$
and originally in the x direction we have $V_{0x}= 12m/s*cos(35)~~=9.8298m/s$

$V_{1y}= V_{0y}-g*Δt~~=4.9209m/s$
$V_{1y}=6.8829m/s -(9.81\frac{m}{s^2}*0.5s)$
$V_{1x}=9.8298m/s$
$V_{1y}= 4.9209m/s$
$\vec{V_1}$ will be constructed from components

and the angle β between from horizontal upwards can be calculated

$tan(β)=\frac{4.9209}{9.8298}$
β=26.5930 deg$sin(26.5930) = \frac{V_1y}{|\vec{V_1}|}~~$
$|\vec{V_1}|=10.9927 m/s$

2. Displacement

this was much harder for me. I could not really remember much useful facts about the displacement formulas etc...

I did seem to grasp that what the teacher wanted us to do was as follows: essentially find the position x_1 and y_1 of the ball. Then you have found the endpoint coordinates. And we know already that the coordinates for the launchpoint x_0=0m and y_0= 0m

displacement is probably the vector from the launchpoint until the endpoint. Displacement is a vector quantity.

I think we know from the endpoint, that this position occured, at the timestamp of 0.2 seconds. (?)

I think the teacher gave us a formula for this type of ball throw scenario to use. But I'm not very confident in knowing and understanding that displacement formula and why it should work the way it does.
Rather than simply engage in a mindless exercise about plugging and chugging values into a formula. I decided to ask about my confusion about that formula. (especially the displacement formula for component form of the displacement vector)

component form equation for velocity :
$V_{x0} = V_{x1}$ (((essentially it seems this is from the fact that there was no x-directed forces and no x-direction acceleration affecting the ball)))

$V_{y1} = V_{y0} - g*Δt$ (((essentially to my mind this formula means that the y component of velocity will be that much smaller velocity. This is because the small g acceleration does affect indeed in the y component direction. Because small g is the acceleration due to gravity and is in the same direction as the Gravity force)))

component form equation for displacement
$X_{1}= X_{0} + \frac{V_{x0}+V_{x1}}{2}* Δt$ I think that the X_1 and X_0 here probably refer to what the x coordinate are for the purpose of calculating eventually displacement. I think X_0 was essentially x value from the origin (x value from launchpoint 0meters)

$Y_1= Y_0 +\frac{V_{y0}+V_{y1}}{2}*Δt$ In similar vein for Y_1 and Y_0, but I'm having a hard time justiffying the formula to myself. Do the two displacement formulas really work llike that , and why should they work like that. It was not immediately obvious to me.

1.)Apparently one source of my confusion is the justification for the [(V_0 + V_1)/2] * Δt.
Actually the entire displacemnt formulas are rather confusing.2.)Why did the formula only have the component values as the fraction terms $\frac{V_{y0}+V_{y1}}{2}$
I think in simpler linear motion problems we just used the regular equation $s= (1/2)*a*t^2$
where does the idea come from that:
$\frac{V_{y0}+V_{y1}}{2}*Δt= (y_1 -y_0)$

3.) I thought that the formula for average velocity was basically as follows $\frac{Δx}{Δt}=v_{avg}$
But our teacher wrote in his notes essentially that "average velocity between t_0 and t_1 time interval is
$\frac{v_0 + v_1}{2}$"
how could both be true for average velocity, unless they really are both identical compared to each other, of course in that case all would be well.
furthermore
$Δx = \frac{v_0 + v_1}{2}*Δt$4.) I have a vague memory that the kinematic equation $s= 0.5* a* t^2$ was somehow related to the idea of taking the average out of the velocities, and multiplying by time.
Our teacher did make a graphical proof of some sort, but I forgot what it looked like sadly. Apparently its a relatively common high school proof type for this equation. Apparently the more complicated justifications usually involve calcuulus principles, which we haven't really done so far.
I know that the kinematic equation deals with calculating total displacement in motion where acceleration is constant value.I think for me the issues 2.) and 3.) are the worst sources of confusion at the moment.
If anybody knows a graphical based proof (geometric proof) for the
$Δx = \frac{v_0 + v_1}{2}*Δt$
I would like to see that picture or video where it is explained.

 

[Simon Bridge]

I think this is the sort of confusion that comes from using formulas without understanding them.
Have a go sketching velocity-time graphs for the motion: you need one for each direction component.
The area under the v(t) line is the displacement, in that direction, and the slope of the line is the acceleration component in that direction.
Since the graphs are lines, the areas are all rectangles and triangles ... so the maths is easy.
The trick is to put variables in where you don't know the numbers.

From examining the graphs, your confusions should clear up somewhat.

 

[Simon Bridge]

I was quite a bit confused conceptually speaking about the total displacement between the time period of 0.2s from the 0s timepoint onward.

1. velocity

according to my notes, it should be so that the $\vec{v_1}$ can be constructed from components.

This is correct ... if you sketch out the initial scenario, the velocity will be drawn as an arrow that is angled 35deg from the horizontal.
Next step is to put axes on the diagram. It is useful to make the +y axis vertical upwards and the +x axis horizontal to the right, starting the ball at x=0, y=0.

it was known in the beginning that $V_{0x}= V_{1x}$ because the only force affecting is in the vertical direction from gravity. Thus I think the x direction is unaffected by a force. y direction does have gravity though.

Correct - so what do the v-t graphs of the motion look like for each direction?
But the basic reasoning that followed [deleted] looks sound.

2. Displacement
this was much harder for me. I could not really remember much useful facts about the displacement formulas etc...

I did seem to grasp that what the teacher wanted us to do was as follows: essentially find the position x_1 and y_1 of the ball. Then you have found the endpoint coordinates. And we know already that the coordinates for the launchpoint x_0=0m and y_0= 0m

displacement is probably the vector from the launchpoint until the endpoint. Displacement is a vector quantity.

This is correct ... technically the displacement is a change in position.
The position vector points from the origin of the coordinate system you choose to a point in that coordinate system, and has length equal to the distance between the origin and that point. It can be thought of as the displacement from the origin. The displacement is the vector showing the change in position ... so an object going from point P at $\vec p$ to point Q at $\vec q$ has undergone a displacement $\vec s = \vec q - \vec p$. ("change in" means final minus initial.) This means that if the object went from P to Q then back to P, the final displacement is zero.

Careful though: the word is often used to mean the displacement from some reference point ... ie the origin. This is the situation here.
The displacement will be the vector $\vec s = x\hat\imath + y\hat\jmath$.

I think we know from the endpoint, that this position occured, at the timestamp of 0.2 seconds. (?)

That is what the problem statement says.
The final time is $t_1$ and the initial time is $t_0$.

I think the teacher gave us a formula for this type of ball throw scenario to use

. But I'm not very confident in knowing and understanding that displacement formula and why it should work the way it does.
Rather than simply engage in a mindless exercise about plugging and chugging values into a formula. I decided to ask about my confusion about that formula. (especially the displacement formula for component form of the displacement vector)

Well done :)

component form equation for velocity :
$V_{x0} = V_{x1}$ (((essentially it seems this is from the fact that there was no x-directed forces and no x-direction acceleration affecting the ball)))

$V_{y1} = V_{y0} - g*Δt$ (((essentially to my mind this formula means that the y component of velocity will be that much smaller velocity. This is because the small g acceleration does affect indeed in the y component direction. Because small g is the acceleration due to gravity and is in the same direction as the Gravity force)))

Here you can write, for an arbitrary time $t$, $\Delta t = t-t_0$ ... since $t_0=0\cdots$ ??

Aside: best to reserve underlining for URLs ... when you type, emphasis is achieved with italics and headings are bold.
You underline headings in handwriting because it's hard to do boldface that way.

Good use of equation rendering ... later on you may want to learn LaTeX, which is standard for science writing and very powerful.

component form equation for displacement
$X_{1}= X_{0} + \frac{V_{x0}+V_{x1}}{2}* Δt$ I think that the X_1 and X_0 here probably refer to what the x coordinate are for the purpose of calculating eventually displacement. I think X_0 was essentially x value from the origin (x value from launchpoint 0meters)

... you seem to be still copying what the teacher has written. Ask yourself "what makes sense?" and write it down.
In physics, maths is a language ... so you think to yourself what is happening in physics and then you convert the words into maths.
In general, an object does not start at the origin ... so you need an initial position and a final position and how the object travels between them.
For this situation, you shuld be using velocity-time diagrams to guide you.

For example:
For the x direction, the velocity time diagram is a horizontal line.
You know this because there are no unbalanced forces in the x direction.
The line goes from $t=0$ to $t=t_1$, and passes through $v_x=v_0\cos\theta$
The displacement is the area under the line ... that shape is a rectangle, so the area is base times height.
The base is $t_1$ and the height is $v_0\cos\theta$ so the displacement is $s_x=v_0t_1\cos\theta$ ... and you are done.
Notice how I did not need to memorize any suvat equations to get there, but you can see that the equation must be correct because it just talks about the area of a rectangle?

For the y direction, there is a constant unbalanced force (gravity) pointing in the -y direction. So you know the graph must be a line sloping downwards.
It starts from $v_y(0)= v_0\sin\theta$ at $t=0$, and ends at $v_y(t_1)=v_1\sin\theta$ at $t=t_1$
The displacement is the area under the line ... that shape is a trapezium, what is the formula for the area of a trapezium?
(If you do not know, divide the area into a rectangle and a triangle and do it that way.)

1.)Apparently one source of my confusion is the justification for the [(V_0 + V_1)/2] * Δt.
Actually the entire displacemnt formulas are rather confusing.

Hopefully they have been explained above ... they are the formulas for the area under the v-t graph between the times of interest. All you need to know how to do is how to sketch these graphs from a description, and remember the formulas for simple areas. later you will learn calculus, which will make the relationships even more plain.

2.)Why did the formula only have the component values as the fraction terms ...

... same answer as above.
It's geometry.

3.) I thought that the formula for average velocity was basically as follows $\frac{Δx}{Δt}=v_{avg}$
But our teacher wrote in his notes essentially that "average velocity between t_0 and t_1 time interval is
$\frac{v_0 + v_1}{2}$"
how could both be true for average velocity, unless they really are both identical compared to each other, of course in that case all would be well.
furthermore
$Δx = \frac{v_0 + v_1}{2}*Δt$

That is true if the acceleration is a constant.

4.) I have a vague memory that the kinematic equation $s= 0.5* a* t^2$ was somehow related to the idea of taking the average out of the velocities, and multiplying by time.

Technically: $s=ut+\frac{1}{2}at^2$, and it's one of those equations you don't need to memorize if you know how to sketch the v-t diagrams.

The above should address your confusions directly. The hole in your understanding is definitely sketching the velocity-time diagram, so this is what you need to learn. I think the Khan academy vids on youtube have a sections on deriving the suvat equations from graphs but this is a place where you have to learn by doing.

 

[Simon Bridge]

In general, an object starts at time $t=0$ with initial velocity $u$ and has velocity $v(t)$ at time $t$ after undergoing constant acceleration $a$.

The graph of this motion is a straight line with slope $a$ and intercept $u$ ... the general equation of a straight line is $y=mx+c$ ... so this line has equation $v=at+u$ ... but I'll procede differently. Start by sketching the graph: draw v vs t axes, draw in points for (0,u) and (t,v), make v > u; draw a straight line through them.

By definition acceleration is the rate of change of velocity. (In maths: $a(t)=v'(t)$)
For constant acceleration, this is the slope of the v-t graph (slope = rise over run): $a=(v-u)/t$
... we like our equations written on one line so: $v=u+at$ is preferred...call this eq(1)

By definition, the displacement is the area under the v-t graph (in maths: $s=\int_0^t v(t')\;dt'$)
a. the area is a trapezium: the formula for a trapezium with parallel sides length a and b separated by width w is $A=\frac{1}{2}(a+b)w$
... so $s=\frac{1}{2}(v+u)t$ ...call this eq(2)
b. the area is a triangle, height v-u and base t, on top of a rectangle of height u and base t:
... so: $s = \frac{1}{2}t(v-u) + ut$. Prove that this is the same as (2)

All the suvat equations can be derived from these two
ie. let's say you don't know the final velocity and you want the displacement:
sub (1) into (2) to get $s=\frac{1}{2}\big((u+at) + u\big)t = \frac{1}{2}(2u+at)t = \frac{1}{2}(2ut+at^2)= ut+\frac{1}{2}at^2$

There are two other suvat equations, see if you can derive them from (1) and (2).

- therefore these are the only two equations you need to know... and you don't even need to know those if you know how to draw the graph, because you have already memorized all the formulae for the areas of shapes that you will ever need.

 

[late347]

This is correct ... technically the displacement is a change in position.
The position vector points from the origin of the coordinate system you choose to a point in that coordinate system, and has length equal to the distance between the origin and that point. It can be thought of as the displacement from the origin. The displacement is the vector showing the change in position ... so an object going from point P at $\vec p$ to point Q at $\vec q$ has undergone a displacement $\vec s = \vec q - \vec p$. ("change in"

Careful though: the word is often used to mean the displacement from some reference point ... ie the origin. This is the situation here.
The displacement will be the vector $\vec s = x\hat\imath + y\hat\jmath$.

I was a little bit confused about that displacement vector notation. But the confusion it seems was not fatal, but just strange...
Our class only really did those simpler displacement type of problem. The definition that the teacher gave us was that displacement is change in position. I guess that I only treated the matter simply as a line segment between the endpoint and the startpoint. I think our teacher mentioned that notation as was above, but I will likely learn it later, because it is not probably in the exam. Was it something like the basis vector? The strange i and j

furthermore why is p vector such as $\vec p$

For example:
For the x direction, the velocity time diagram is a horizontal line.
You know this because there are no unbalanced forces in the x direction.
The line goes from $t=0$ to $t=t_1$, and passes through $v_x=v_0\cos\theta$
The displacement is the area under the line ... that shape is a rectangle, so the area is base times height.
The base is $t_1$ and the height is $v_0\cos\theta$ so the displacement is $s_x=v_0t_1\cos\theta$ ... and you are done.
Notice how I did not need to memorize any suvat equations to get there, but you can see that the equation must be correct because it just talks about the area of a rectangle?

For the y direction, there is a constant unbalanced force (gravity) pointing in the -y direction. So you know the graph must be a line sloping downwards.
It starts from $v_y(0)= v_0\sin\theta$ at $t=0$, and ends at $v_y(t_1)=v_1\sin\theta$ at $t=t_1$
The displacement is the area
That is true if the acceleration is a constant.

why do I need to do the graphs separately for x and y component each? maybe that's a dumb question but anyway.

Technically: $s=ut+\frac{1}{2}at^2$, and it's one of those equations you don't need to memorize if you know how to sketch the v-t diagrams.

The above should address your confusions directly. The hole in your understanding is definitely sketching the velocity-time diagram, so this is what you need to learn. I think the Khan academy vids on youtube have a sections on deriving the suvat equations from graphs but this is a place where you have to learn by doing.

For the y direction, there is a constant unbalanced force (gravity) pointing in the -y direction. So you know the graph must be a line sloping downwards.
It starts from $v_y(0)= v_0\sin\theta$ at $t=0$, and ends at $v_y(t_1)=v_1\sin\theta$ at $t=t_1$
The displacement is the area under the line ... that shape is a trapezium, what is the formula for the area of a trapezium?
(If you do not know, divide the area into a rectangle and a triangle and do it that way.)

Hopefully they have been explained above ... they are the formulas for the area under the v-t graph between the times of interest. All you need to know how to do is how to sketch these graphs from a description, and remember the formulas for simple areas. later you will learn calculus, which will make the relationships even more plain.

... same answer as above.
It's geometry.

I think the picture about the displacement as area under the line makes more sense in triangle and a box.

I did do another problem with graphing like that with the same strategy. The scenario in that other problem was a straightforward situation of a car braking scenario with the car's velocities written out together with time.

The objective was to find the total displacement in the time interval from 2s -->4s. I think this was easy for me because the car is only moving straight forward even though it is braking at the same time.

the values which existed were tabulated as

it comes out such that we can calculate each displacement which was during one second. And add those displacements together.
I think it comes out as a graphical integration when you put the velocity into y-axis and time into x-axis.

The catch of that exercise was to notice that acceleratino was not constant across that interval because the line is a little bit crooked for velocity. So appparently you have to essentially calculate like two different trapezium and their area together forms the displacement. It looks like the answer to that was about 22.78m

 

[late347]

In general, an object starts at time $t=0$ with initial velocity $u$ and has velocity $v(t)$ at time $t$ after undergoing constant acceleration $a$.

The graph of this motion is a straight line with slope $a$ and intercept $u$ ... the general equation of a straight line is $y=mx+c$ ... so this line has equation $v=at+u$ ... but I'll procede differently. Start by sketching the graph: draw v vs t axes, draw in points for (0,u) and (t,v), make v > u; draw a straight line through them.

By definition acceleration is the rate of change of velocity. (In maths: $a(t)=v'(t)$)
For constant acceleration, this is the slope of the v-t graph (slope = rise over run): $a=(v-u)/t$
... we like our equations written on one line so: $v=u+at$ is preferred...call this eq(1)

By definition, the displacement is the area under the v-t graph (in maths: $s=\int_0^t v(t')\;dt'$)
a. the area is a trapezium: the formula for a trapezium with parallel sides length a and b separated by width w is $A=\frac{1}{2}(a+b)w$
... so $s=\frac{1}{2}(v+u)t$ ...call this eq(2)
b. the area is a triangle, height v-u and base t, on top of a rectangle of height u and base t:
... so: $s = \frac{1}{2}t(v-u) + ut$. Prove that this is the same as (2)

All the suvat equations can be derived from these two
ie. let's say you don't know the final velocity and you want the displacement:
sub (1) into (2) to get $s=\frac{1}{2}\big((u+at) + u\big)t = \frac{1}{2}(2u+at)t = \frac{1}{2}(2ut+at^2)= ut+\frac{1}{2}at^2$

There are two other suvat equations, see if you can derive them from (1) and (2).

- therefore these are the only two equations you need to know... and you don't even need to know those if you know how to draw the graph, because you have already memorized all the formulae for the areas of shapes that you will ever need.

Lets try


I suppose enough proof will be to work from triangular definition and work the example of the box area, plus the triangle area. into the trapezium formula. ?

I will first go from trapezium variables and try to see if that works also for the physics variables like $V_0$ and $V_1$ $t$etc...

$\frac{1}{2}*(a+b)*w$

triangle area is good and clear $\frac{(a-b)*w}{2}$
box area is good and clear $b*w$

expand box fraction
$\frac{bw}{1}+\frac{(a-b)*w}{2}$
$\frac{2bw}{2}+\frac{(a-b)*w}{2}$
$\frac{2bw+aw-bw}{2}$

factorize
$\frac{aw+bw}{2}$
$\frac{(a+b)*w}{2}$
$\frac{(a+b)}{2}*\frac{w}{1}$
$\frac{1}{2}*(a+b)*w$for the displacement equation with physics variables. Unless I have made grave error in interpreting your variables. I will write the variables equally validly as follows.

furthermore it looks like your equation for the displacement in this case is turned around image of my trapezium as follows

s=distance
V_0=starting velocity
V_1= end velocity
t=time

$s = \frac{1}{2}t(v_{1}-v_{0}) + v_{0}t$I proved the above picture case with physics variables essentiially on pen-and-paper, I'm afraid I was too bored and frustrated at writing the equations on computer

 

[Simon Bridge]

I was a little bit confused about that displacement vector notation. But the confusion it seems was not fatal, but just strange...
Our class only really did those simpler displacement type of problem. The definition that the teacher gave us was that displacement is change in position. I guess that I only treated the matter simply as a line segment between the endpoint and the startpoint. I think our teacher mentioned that notation as was above, but I will likely learn it later, because it is not probably in the exam. Was it something like the basis vector? The strange i and j

That is correct ... $\bf{i}$ = $\hat\imath$ is a vector length 1 that points in the +x direction, sometimes written $\hat x$.

furthermore why is p vector such as $\vec p$

The arrow over the top of a letter is one standard notation in the SI system to show that the letter stands for a vector. Many typed materials often use bold-face p and you may have been taught to put a tilda undernieth the letter in handwriting. One advantage of the arrow over the top is that you can type and write the same thing.

Without the arrow, or other mark, it would just be a scalar.
In fact, by convention: $p=|\vec p|$ ... which is to say that the magnitude of $\vec p$ is $p$.
The special "hat" notation $\hat p$ would indicate a unit vector pointing in the same direction as $\vec p$ so I can also write $\vec p = p\hat p$ if I want to be clever.

why do I need to do the graphs separately for x and y component each?

This is because it makes the maths easier ... it is usually the case for motion in more than one dimension, that the maths is likely to be simpler if the dimensions should be graphed separately.
It's basically the same reason you have different equations for the x and y components of things.

I think the picture about the displacement as area under the line makes more sense in triangle and a box.

That is the first way I learned to see it ... the maths works out the same. In exams, the v-t graphs usually work out to be just triangles so it's even easier. Like with the car breaking example you did before:

I did do another problem with graphing like that with the same strategy. The scenario in that other problem was a straightforward situation of a car braking scenario with the car's velocities written out together with time.

The objective was to find the total displacement in the time interval from 2s -->4s. I think this was easy for me because the car is only moving straight forward even though it is braking at the same time.

That is correct ... there was only motion in 1 direction, so you choose that direction for the +x axis and go from there.

...it comes out such that we can calculate each displacement which was during one second. And add those displacements together.
I think it comes out as a graphical integration when you put the velocity into y-axis and time into x-axis.

Finding the area under a graph is, indeed, integration... but we don't like to do calculus if we can avoid it - like when the shapes are simple.
However, try to get out of thinking of x and y axes when you are not using them for x and y. They are "horizontal" and "vertical" axes until they are assigned to something, and when the vertical is velocity then it becomes "the velocity axis" ... but there is no special reason to assign a particular thing to a particular axis besides convenience. I sometimes make say "y=<distance fallen>" then the vertical axis will actually indicate distance downwards and get labelled "y". There are also situations where it may be more convenient to plot y values on the horizontal axis (common in relativity where time is on the vertical and distance is on the horizontal).

The catch of that exercise was to notice that acceleratino was not constant across that interval because the line is a little bit crooked for velocity. So appparently you have to essentially calculate like two different trapezium and their area together forms the displacement. It looks like the answer to that was about 22.78m

... that is the correct reasoning.

So now you can see how the suvat equations come from the v-t graph and the formulas for area.

 

[late347]

So, essentially you are arguing such that:

when we compute the vt graph for the V_x1 component velocity and the time that we want to measure displacement, we are not actually measuring "pure displacement" but we are essentially measuring the x-axis component of the displacement vector which we originally have to calculate.?

What other purpose would be then have to break up into components?

we were only interested in the final displacement vector which will have some upward angle, and some magnitude. The end point of that displacement vector will to my reckoning be some point, at the ballistic trajectory of the ball. But the displacement vector itself will be essentially the shortest distance betwen the endpoint and the startpoint.

 

[Simon Bridge]

.....
furthermore it looks like your equation for the displacement in this case is turned around image of my trapezium as follows...

Yep - in the general example I said that v>u, so the trapezium top has an upwards slope.
For the equation where v<u, you would write: $s=\frac{1}{2}(u-v)t +vt$
You can also start with the big rectangle and subtract the upper-right triangle :) $s=ut - \frac{1}{2}t(u-v)$
... it doesn't matter because the equations end up the same.

s=distance
V_0=starting velocity
V_1= end velocity
t=time

$s = \frac{1}{2}t(v_{1}-v_{0}) + v_{0}t$

Adopting your notation: if you want $v_0 > v_1$ then go
$s=\frac{1}{2}(v_0-v_1)t + v_1t = \frac{1}{2}v_0t-\frac{1}{2}v_1t + v_1t = \frac{1}{2}v_0t+\frac{1}{2}v_1t = \frac{1}{2}(v_0+v_1)t$
... which gets you the same equation.

I proved the above picture case with physics variables essentiially on pen-and-paper, I'm afraid I was too bored and frustrated at writing the equations on computer

I type them in directly, and use copy and paste when making small changes, it's less frustrating that way.
... are you using the built-in equation editor?

 

[late347]

I type them in directly, and use copy and paste when making small changes, it's less frustrating that way.
... are you using the built-in equation editor?

yes but the website is laggy if I want to doublecheck the code in preview mode.

 

[Simon Bridge]

I'm a little behind you so bear with me...

So, essentially you are arguing such that:

when we compute the vt graph for the V_x1 component velocity and the time that we want to measure displacement, we are not actually measuring "pure displacement" but we are essentially measuring the x-axis component of the displacement vector which we originally have to calculate.?

The area under the graph of v_x vs t is the magnitude of the x component of the displacement vector, yes.

Since you know about integration: the x displacement between $t_0$ and $t_1$ is $$s_x=\int_{t_0}^{t_1}v_x(t)\; dt$$

What other purpose would be then have to break up into components?

That's it - it makes the maths simpler.
An example of not breaking the components up may be if a train runs along a track ... we may only want to know about how far the train travels along the track rather than it's x-y-z position on the terrain.

we were only interested in the final displacement vector

which will have some upward angle, and some magnitude. The end point of that displacement vector will to my reckoning be some point, at the ballistic trajectory of the ball. But the displacement vector itself will be essentially the shortest distance betwen the endpoint and the startpoint.

In this case the final displacement = the displacement at t=t1 from the position at t=t0
Since t0=0 and the position at t=0 is the origin, then displacement vector and the position vector are the same.
That won't always be the case.

Of course any vector will have a magnitude and an angle...
$\vec s_1 = s_{1x}\hat\imath + s_{1y}\hat\jmath$ is the representation in cartesian coordinates.
$s_1=|\vec s_1|=\sqrt{s_{1x}^2+s_{1y}^2},\qquad \theta_{s1} = \tan^{-1}(s_{1y}/s_{1x})$ gives you the "polar coordinates".

BTW: helps typing if you use fewer subscripts.

 

[late347]

I did some pen and paper calculations and arrived at some angle and magnitude for the displacement vector of the ball. The endpoint of the ball, the position will be in the timestamp of 0.2 seconds I think.

in the horixontal displacement component for displacement we can get that the displacement component with horizontal velocity component and time (9.8298m/s) * (0.2s) = 1.96596m

For the vertical direction displacement component...
One can draw a vt graph again

displacement component should be the area in green.

Once all the components are gathered around. Then one can draw the displacement vector from its components for instance and calculate the angle and magnitude?

I got angle γ of displacement=30.98 deg
magnitude of displacement =2.293m

 

[late347]

I'm a little behind you so bear with me...

The area under the graph of v_x vs t is the magnitude of the x component of the displacement vector, yes.That's it - it makes the maths simpler.
An example of not breaking the components up may be if a train runs along a track .

Good to know that the Vx and t graph gives out the horizontal displacement component.
I had some trouble justifying the horizontal component to myself though...

I think the vertical component is easier to justify to myself because the acceleration due to gravity cuts into the vertical component as the ball flies.

In real world there would definitely be air resistance for the ball so I think the vertical and horizontal components will not be so simple.

I would imagine that a ball thrown in real life with air resistance at the same angle and velocity... will travel much less distance over flat ground compared to hypothetical situatuin where the ball suffers no air resistance at all...

Did the answer seem reasonable for displacement between the time 0-0.2 seconds.

Displacement from X_0 =0
Y_0= 0
Angle ~ 31 deg above horizontal
Magnitude ~ 2.3m

 

[late347]

Displacement during 0s --> 0.2s