SPH3U - Kinematics - Projectiles 1

okay and here we are um lesson 10 continuing our discussion of motion in two dimensions and we looked at relative velocities

um and kind of adding velocity vectors and now we're going to investigate something called projectile motion which is the motion of objects as they

move in two-dimensional space and until this point we've we've really looked at the motion of

objects would travel in one dimension so cars that are either accelerating or moving um kind of left or right or north or south or east or west or up or down

okay but through the use of vectors what we're going to do is extend our study of motion now into two dimensions so looking at things that move left and

right and up and down and do so simultaneously so let's begin by discussing parabolic

motion or projectile motion and we're going to look at two different cases let's say you're out in the middle of in in the middle of space so there's you

know there's nothing around you there's no planets there's nothing you're in the middle of empty space so there's no gravity anywhere and if you were to throw a baseball it will leave your hand and travel in a

straight line forever and so the position vector of the ball or the motion of the ball would be just a straight line if you're in outer space

but on earth we know that that doesn't happen when we throw a ball on earth it doesn't travel in a straight line

it travels in an arc it kind of falls to the ground and that's a major difference

and of course that's because of gravity and we know that it's because of gravity but the term projectile motion is really used to describe events where an object moves under the influence of

gravity only and is not self-powered okay so um rockets or anything like that really don't travel in projectile motion

because those things are self-powered but a baseball for example travels only under the influence of gravity as soon as it leaves the thrower's hand so

let's take a look at the motion this kind of projectile motion and and look at it very specifically because we can make a couple of observations the first observation that we make

is that the ball does not change its horizontal velocity okay so we're neglecting air resistance but virtually in the x direction in the horizontal direction

there's no acceleration once it leaves the pitcher's hand it travels with the same horizontal velocity

indefinitely we do know though that the velocity in the downward direction so in the vertical direction gets larger due to the acceleration of

gravity because gravity accelerates things as we know towards the earth and the trajectory of the ball would kind of look like this and so

the blue line kind of represents the resultant velocity but we can draw the pieces that make up that resultant velocity so here's all the horizontal velocities

so from position 1 to position 2 position 2 to 3 position 3 to 4 position 4 to 5 the velocities in the x direction are

the same but what do we see with the velocities in the y direction well they start getting larger in the

downward direction because we have an acceleration from gravity that's downwards

what we want to notice is that each horizontal component is constant but each vertical component of velocity increases and of course we've said

before that's due to gravity we also know that the angle that we launch a ball or a projectile or really anything in general also affects the path

because you know that if you throw a ball think about baseball if you throw a ball without very much of an angle it doesn't go that far if you throw a ball

in a pop fly like right up in the air it doesn't go that far the range which is the horizontal displacement in x is greatly affected by the launch angle

and what you'd actually find is that in a launch angle of 45 degrees so right halfway in between the x and the y direction you get maximum range and that's

something that we'll consider a little bit later and here's two diagrams to kind of illustrate that

here's a thought experiment let's say you and a friend each have a tennis ball okay at the exact same moment and at the exact same height above the ground you and your

friend both release your tennis ball you throw your ball perfectly horizontal so you actually throw it with a horizontal velocity perfectly

horizontal your friend simply releases their ball from rest okay so you guys are standing at the exact same spot

you throw it horizontally your friend doesn't do anything they just release it and drop it to the ground okay they release it from rest whose

ball hits the ground first and of course in class we talk about this and we kind of ask for thought so you know think about what you think is going to

happen and there's a couple of things that could happen well you throw your ball horizontally maybe your ball hits the ground first

but you may be thinking well it travels a greater distance which of course it would so maybe your friend's ball hits the ground first so what's the answer the answer is of

course if we neglect air resistance both balls hit the ground at the same time there's no acceleration in the x direction

and so even with an initial x velocity the time to hit the ground is the same and of course the question is why why is the ground to hit why is the time to hit the ground the

same and if you think about the distance between your ball and the ground and the distance between your friends ball in the ground is the same and because the acceleration

of the two balls is the same it's both they're both being pulled to the ground in uh by gravity at a certain acceleration

that's the same they both have no initial velocity in the downward direction because you released your ball perfectly horizontal well then they both must hit the ground

at the exact same time so think about the equations of kinematics that you'd use to show that let's take a look at an example so we can actually start using some of these

equations a baseball is thrown vertically at 35 kilometers per hour excuse me baseball's thrown horizontally at 35 kilometers per hour

let's calculate the velocities in the horizontal and the vertical direction at one second and at four seconds and then what's the final velocity at

each of these times okay so here's our little diagram we got the baseball at the initial spot you can see there one second later it's traveled a certain

distance and then four seconds later it's traveled a farther distance let's calculate horizontal and vertical velocities

for each of these positions and we'll work on the final velocity so we know that initially v1

is 35 kilometers per hour but it's horizontal because we say that the ball is thrown horizontally so you should be thinking oh that's a velocity in the x direction

there is no velocity in the y direction at this time and there we have t equals one second and t equals four seconds so now we're kind of on the same page we get everything let's first because we're talking about

seconds and we've got a velocity that's in kilometers per hour let's convert that right away 35 kilometers per hour

there's a thousand meters per kilometer 60 seconds in a minute and 60 minutes in an hour so if we do the conversion 9.72 meters per second east

all right so first of all let's just say this in general what do we know about horizontal velocities because this is going to tell us with a projectile

the horizontal velocities always are the same and you should be saying that oh yeah that's right because we know that with a projectile

where's the acceleration the acceleration is down because gravity is pulling the ball down there is no acceleration in the horizontal direction so it means

oh okay so the x velocity is going to be the same at the beginning at one second at two seconds at three seconds at four seconds at ten seconds the x velocity the

horizontal velocity is always the same so let's divide the remainder of this into kind of three separate sections here okay so here's our three little separate sections this is going

to help us to organize our solution and so let's look first at the horizontal velocities because they're the easiest because we know they don't change at t equals zero

our horizontal velocity is 9.72 meters per second at t equals 1 our horizontal velocity is 9.72 meters per second

at t equals 4 our velocity is 9.72 meters per second so this is one of the important things about projectile motion is that the horizontal velocity of the projectile

never changes it's always the same and so there it is now though we know that the ball is being accelerated in the downward in the vertical

direction so let's take a look at now what the vertical velocities are going to be well at t equals zero the velocity in the y

direction is zero because it's launched at 35 kilometers per hour yeah yeah but that is a horizontal velocity so what we're doing here is we're

actually splitting the question into two parts one part looks at horizontal velocities and one part looks at vertical velocities and this

method of splitting the problem into horizontal and vertical directions is going to be an approach that you'll take to solve

all kinds of projectile questions later so we know the initial y velocity of this specific

projectile is zero let's find out what it is then at one second okay so there's a little arrow to just remind

you why that vertical velocity is zero so here's t equals one and let's use one of our kinematics equations v two equals v one plus a t

but this is in the vertical direction okay so what's v1 in the vertical direction zero what's the acceleration in the vertical direction well it would be 9.8 meters per second

squared that's the velocity due to gravity or sorry excuse me the acceleration due to gravity and it's in the downward direction so we'll give it a negative

and of course the time is one second so after one second it's gone from zero to nine point eight meters per second and it's in the downwards direction

so there's v2 at time t equals one what about at time t equals four well let's do the same thing we want to know v2 so really the second

velocity but at four seconds v one we can still say is zero but now instead of subbing in one for time now we're at four seconds

so the downward velocity at 4 seconds is 39.2 meters per second and so what we do is we say okay so now we know what our x component of velocities are and now we know what our y

component of velocities are and so really we've answered the first part of this question calculate the velocities in the horizontal and vertical directions now what we want to do is we want to say

okay so then how does the overall velocity change and this is where our use of vectors comes into play because just like we did with displacements and

stuff like that we now have a horizontal velocity and we have a vertical velocity and we can put these two things together using vectors to find a total and we can

find an angle and so here's a little diagram of t equals one second

so we have a horizontal velocity of 9.72 and a vertical velocity downwards of 9.8 and so we can use the pythagorean

theorem to find the final velocity and it ends up being 13.8 meters per second we can also find the angle

that this velocity would be at tan theta equals 9.8 over 9.72 and if we take the inverse tan of that ratio

we find that the angle theta is 45.2 degrees okay so we would report this then because it's a vector we need to report

the direction so we'd say that the final velocity is equal to 13.8 meters per second and it's in a direction of east 45

degrees down or you could say east 45 degrees south or you could say right 45 degrees down any of these

combinations is just fine so on your own try the last part of this problem which is to find the total velocity the final velocity

at four seconds try that on your own using the exact same method what would you expect you'd expect to be a greater velocity in v final v final would be would be higher

than 13.8 and you'd expect that it would be at an angle greater than 45 degrees okay

here's example two an arrow is shot from a height of 20 meters with an initial horizontal velocity

of 18.0 meters per second how far in the horizontal direction will the arrow fall to the ground

another way of asking this question is what's the range of this arrow so now we're actually looking for

not a velocity but a displacement because how far indicates a displacement so here we have a little diagram we got you know somebody standing on i don't know a 20 meter high cliff and

they fire this arrow and of course the arrow travels in parabolic motion until it hits the ground so again um first of all you know before we do anything we should say okay yeah

we're going to say up as positive as per our convention and we're going to say that east is positive or right is positive as per our convention

okay so that's the first thing let's take a look at what we know we know v1 in the x direction and we're going to call that v1x

right so we know v1 and we know it's in the x direction and it's 18 meters per second we also know v1

y and it's not listed here but of course what is v1y and you should be saying to yourself oh v1y is

zero because it's launched with a horizontal velocity we know that the arrow basically we're looking for the distance that it's going

to take for the arrow to hit the ground so we know that it starts d1 it's for its initial displacement in the y direction

is up in the air 20 meters and what's its final displacement d2 in the y direction we know it's zero because we're looking for the point at which it hits the ground and we're saying we're calling

the ground zero we also know the acceleration in the y direction a y and it's down 9.8 meters per second squared so that's everything that we know

how long will it take for the arrow to hit the ground because before we can figure out how far the arrow is going to go we need to figure out how long it's going to take to get there and this

might be something that's useful so let's investigate well let's take a look at equation number three kinematics equation number three but only in the y direction and so this is what i'm doing i'm kind of

separating the problem out into the horizontal components and the vertical components because it's important to know that we don't ever want the horizontal

and the vertical to come into contact with each other they can't uh an x can't make its way into a y equation and a y can't make its way into the

x equation all right it's like the it's like romeo and juliet right the montagues and the capulets they you know they can't uh never can they mix and that's

the same with x's and y's all right so how long will it take the error to hit the ground well let's solve for t we know that the change we know the change in displacement in the y

direction okay we know that it's negative 20 because it starts it goes from it goes from 20 to zero so it's negative 20. the arrow is going to fall

20. the arrow is going to fall 20 meters we know that the initial velocity in the y direction is zero so that's going to wipe out that term altogether

we know that the acceleration is down 9.8 meters per second squared so this equation now actually becomes pretty straightforward we can solve this no problem because this is just a simple little

quadratic and what we find is is that of course t squared is equal to 20 divided by 4.9 and when we take the square root of that

of course we get plus or minus 2.02 seconds but because in general the type of physics that we're working with a negative time doesn't make a lot of sense we actually don't need the

negative root so what this has just told us is is that hey this arrow falls to the ground

in 2.02 seconds okay that's good because now we have this other piece of information in its time and unlike the x's and the y's which

have to stay with the x equations and the y's stay with the y equations time can move back and forth right

time is a friend to both sides and so time can kind of show up in the x equations and it can also show up in the y equations so now if we know that it takes the

arrow 2.02 seconds to go from 20 meters to zero meters we can now find out how long it takes for the error to travel the same distance but in the x direction

okay so how far will it go in x in 2.02 seconds well here's again equation number three but is there an acceleration

in the x direction does the arrow accelerate in the x direction at this point you should be shaking your head and you should say no the arrow does not accelerate it's a

projectile projectiles don't accelerate in the x direction they only accelerate in the y direction so that

term that final term there the one half axt squared becomes zero we know that the arrow starts out

at a horizontal displacement of zero and we know that it goes some distance d2x so all we need to do is multiply

our 2.2 seconds 2.02 seconds times the 18 meters per second and we find that the ball that the arrow travels 36.4 meters

in the x direction and that's it more uh projectile motion questions tomorrow