Understanding Projectile Motion: A Comprehensive Guide

the tying run is that second the run that would win the world series is at first and joe carter is the batter

now the two good morning today we'll be investigating projectile motion

and we're going to start off with a baseball question a ball has a speed of 108 kilometers an

hour 35 degree angle with respect to the horizontal

hit one meter above the ground and it's hit by a bat a person is patrolling the outfield

approximately 60 meters from where the ball was hit and the person expects to make the catch

with his glove placed about one meter above the ground the ball is going to land somewhere behind

the person's initial starting position how far does the person have to move to make the catch

so here's the basic diagram the ball is going to be hit at a specific angle it's going to land and ultimately we're

solving for the range or the distance that this ball is going to travel

this is a classic physics problem so let's start off by dissecting this question

a ball has a speed of 108 kilometers per hour 35 degree angle and so what does that look like well we

draw the velocity vector we label it 108 kilometers an hour or 30 meters per second remember to

convert from kilometers per hour to meters per second we divide by 3.6

now we draw an x and a y direction why do we do this well in the x direction we're assuming no forces

in the y direction we're assuming there's only one force the force of gravity

and we break apart the velocity vector into its x and y components

we label the 35 degree angle there's v x and there's v1 y

so why the different subscripts this is such a small point in physics but why did i not say

v1x why did i just write down vx why was i so specific by saying v 1 y

well we know in the x direction the velocity is constant so there is no point to writing v1 or v2

the velocity doesn't change in the y direction however the velocity does change it changes because of

gravity the ball when it's going upwards it's going to slow down

then when it starts to fall downwards it's going to speed up clearly in the y direction the velocity

changes that's why we need to use the one subscript meaning

initial velocity in the y direction all right pause the video right now please use some trigonometry to solve

for vx and v1y hopefully you've tried this already to

get the x we're going to use the cosine ratio being adjacent

over hypotenuse cosine 35 degrees is vx which is the adjacent over 30 meters per second which is the

hypotenuse cross multiplying we end up with this value for vx

and there it is for the y component of the vector v one y

we're going to use the sine ratio sine 35 degrees is opposite over hypotenuse v1y over 30

cross multiplying and there's v1y and there it is labeled next the ball is hit one meter above the

ground and then it's caught one meter above the ground

and so that's what that looks like one meter it's hit above the ground and it's caught one meter above the

ground and that's the question how far does the person have to move to make the catch

so that's our goal to get that distance there where i've labeled it with the question mark

and so to do this we have to get distance in the x direction or as mentioned at the very beginning of

the question we have to get the range and so starting off with our x direction we know in the x direction it's constant

speed that's because there are no forces or we are assuming any ways that there are no forces acting in the x

direction we can use this simple equation and substituting we notice there's a

problem we don't know the time often in these projectile motion questions

we have to use both directions to solve the ultimate goal so far we haven't used the y direction

so let's see what we can learn from the y direction and so we note the y direction here we

know the v1 17.207 meters per second in the upwards direction

here's our acceleration and our displacement notice our displacement

is zero is the displacement zero well at the very beginning the ball is hit

one meter above the ground at the very end of the trip for this ball

the ball is caught one meter above the ground so it's hit one meter above the ground

it's caught one meter above the ground it's displacement in the y direction is zero

its final position and its initial position the y direction are the same

so its displacement is zero all right there is our equation we are going to use

please pause the video and solve that equation now we're solving for time

okay i hope you work through the math substituting the displacement the initial velocity and the

acceleration multiplying half of 9.8 rearranging the equation

cancelling out one of the times and dividing by 4.9 we end up with this time

what does this time mean it means that right after the ball was hit the ball will stay in the air for exactly 3.511

seconds before it's caught by the person in the outfield that's what that time means it's the

time of flight and so now we're going to substitute that time back into the equation

for the x direction and solve for distance and so substituting and cross

multiplying we end up with 86.3 meters and so this ball will travel 86.3 meters

from where it was originally hit taking the difference of 86.3 meters and

60 meters the person will have to travel 26.3 meters to catch

the ball our next example a projectile is launched from a point on level ground

and has a speed of 3 meters per second at an angle of 45 degrees to the horizontal which of the following

statements is not true so this is a multiple choice question

which of the following statements is not true a horizontal component of velocity is

constant b at maximum height the vertical component velocity is zero

c the path of the projectile is parabolic or d the initial horizontal component of

velocity and the initial vertical component velocity are equal to half

the initial velocity please pause the video and give this some thought okay i hope you've tried this

now before we get to the answer i want to discuss this part c is the path of the projectile

parabolic let's see and so analyzing this video extracting frames

every 0.1 seconds we can recreate this image and clearly it looks like it's a

parabola i think we could find a more exciting demonstration

showing that in fact projectile motion is parabolic when there is no drag let's see

does it go in i think we all know the answer to that wonderful parabolic motion and so yes

based on the two examples you just saw the path of the projectile is clearly parabolic when there is no drag

and so the answer is d the initial horizontal component of velocity and the initial vertical component of velocity

are equal to half the initial velocity this is not true why is that well here's the vector and i'd like you

to show this i'd like you to show this as an example you're going to see in fact that it's

not half the value and our final example today a cannon makes an angle

greater than 20 degrees with the horizontal and is placed on top of a cliff which is

25 meters high the cannonball is launched with an unknown velocity

we're going to assume the acceleration due to gravity is negative 10 not 9.8 just to simplify the math and lens

173.205 meters from the base of the cliff five thousand milliseconds later

calculate the components of the initial launch velocity the launch speed and launch

angle let's focus on a so this is the diagram that we have and the question is what is that launch

velocity so we need to break this apart once again into an x and a y

direction and we need to label the vx and the v1y for this vector so step one we're going to solve for vx

step 2 we're going to solve for v1y and step three once we have v x and v one y pythagorean theorem will give us

v for the speed please pause the video now and see how far you can get with this question

all right vx well we need to know the distance the distance was listed as 173.205

meters and for the x direction we can use a simple formula

why is that there is no acceleration in the x direction why is that there is an assumption of no

forces in other words no drag acting in the x direction

remember the ball takes five seconds to land and that's why we're using five seconds in our equation

step two solve for v one y well our displacement is 25 meters how do we know that well the ball drops

ultimately 25 meters below from where it started there's the mathematics involved and

there's v1y 20 meters per second so labeling that in our triangle

step 3 solve for speed there's our triangle here's pythagorean theorem and there's

our answer 40 meters per second the launch angle we're going to use

opposite over adjacent which is the tangent opposite is 20 adjacent 34.641

and there's our launch angle 30 degrees find the final components of velocity of the cannibal

final speed and final angle just before it hits the ground please review the description of this

video for that answer for b and finally c find the maximum height of the cannonball

well to get maximum height we have to look at information in the y direction and that's what we're looking for

that distance drawn there remember at maximum height the velocity in the y

direction is zero this is very important that you note that still moving in the x direction but in

the y direction it's no longer moving upwards or downwards at peak height all right pause the video right now and

see how far you can get with this all right so let's see let's see our variables

well there they are we know v2 y is zero there's v one line acceleration we're looking for the displacement

there's the equation we're going to use the classic equation of physics and there's the answer for displacement

in the y direction or maximum height there it is listed and so i hope you've enjoyed part two of

projectile motion if you haven't already seen part one it'll be very helpful to do that

have a great day bye