Understanding Projectile Motion: Forces, Velocity, and Calculations

good morning today we are going to investigate projectile motion by using

this simulation the first thing i want to show you are the forces

acting on the projectile as it's moving there is only one force that we will assume is acting on the projectile and

that's gravity for now we will assume that there is no air

resistance before we continue with the simulation i want you to notice a few things

in the upper right hand corner notice i have broken up the motion into an x and a y direction

why have we done this well as you just saw we're assuming there's no drag

so we're assuming there's no forces in the x direction and the only other force is gravity and

that's acting along the y direction so that's why we're able to break

up this problem into an x and a y direction there is a row of data that you'll

notice that's related to the x direction 228.1 meters refers to the distance

the object has traveled from where it started also notice that 29.6 meters per second

refers to the current velocity in the x direction and now i really want you to pay

attention to that specific number and only that specific number when we launch the projectile again so we won't

change any parameters and i just want you to focus on current velocity in the x direction

let's see notice that the current velocity in the x direction

does not change throughout the journey let's change the angle and see if that holds true

at a different angle notice the current velocity in the x direction has changed but throughout the journey it remains

constant this time the current velocity in the x direction was only 17.2 meters per

second and that number did not change so i want you to keep that in mind

now let's pay attention to the y direction this row of information gives us some

details about the y direction it tells us the maximum height of the object

in addition it also tells us the current velocity in the y direction now based on this diagram this

represents the velocity just before the impact it's negative because the negative means

that the object was moving down so just before impact the object was moving downwards

let's focus on current velocity in the y-direction notice it's positive while it's going

upwards and negative while it's going downwards let's focus in on current velocity in

the y direction when the object hits peak height right around now

notice that the current velocity at peak height in the y direction was zero

this will be easier to see if we use a greater angle right around now

and paying attention to current velocity in the x direction notice that again x direction velocity does not

change and at peak height the y velocity is zero keep this in mind today

so in summary before we continue throughout the journey velocity x remains constant

and at maximum height velocity y is zero the force is acting on a baseball or cannonball or any projectile as it's in

flight are gravity air resistance and possibly lift

however for all of these problems we will assume that the only force acting on the projectile is gravity

hence acceleration or changing velocities only takes place in the vertical

direction in the y direction and the velocity in the horizontal direction in the x direction

remains constant when these assumptions are made the motion that is modeled is parabolic

in nature here's our first problem a cannonball is fired from the top of a cliff

the initial speed of the ball is 72 kilometers an hour find the distance the cannonball will

land from the base of the cliff and calculate the speed of the cannonball 2500 milliseconds

after launch so this is the picture we sort of have we have a cliff 50 meters tall

we have a cannon at the top of it and we're launching a projectile in the manner you see at 72 kilometers

an hour and so the question is what's the distance

we're going to break this up into an x and a y problem now for the x direction we know the

speed is constant 72 kilometers an hour in meters per second that's 20 meters per second we divide by 3.6

because the velocity in the x direction does not change we could use this formula

speed equals distance over time substituting we know that our speed is 20 meters per

second but the problem is we don't know the distance and we don't know the time

so often in projectile motion questions we need to use both directions to solve a problem

in the y direction we know that the cannonball is going to drop 50 meters so its displacement is 50 meters down

we also know the acceleration it's 9.8 meters per second per second that's down

we also know that our initial velocity in the y direction is zero how do we know that if we pay very

careful attention to our initial velocity notice that the vector is pointing along

the x direction that's 72 kilometers an hour that 20 meters per second is pointing in the x

direction only it's not pointing upwards it's not pointing downwards it's pointing only in

the x direction it lines up perfectly with the x direction so this is the reason why

the initial velocity in the y direction is zero the cannonball was aimed upwards or

downwards this wouldn't be the case it's only because the cannonball is aimed

along the x direction here's our displacement formula please now pause the video and try to

solve for time i hope you've tried this substituting our displacement

and our acceleration continue on with the math multiplying half of 9.8

dividing by 4.9 we end up with this answer for time squared

and here's our time this is telling us that the projectile is in the air for 3.194 seconds before

hitting the ground so from when it's launched to when it hits the ground that's the time it takes

for this to happen now going back to our x direction we are going to take 3.194 seconds and

substitute it in for time there's a substitution and working through the math we end up with this

answer for distance significant digits it's 60 meters why only one significant digit

while the 50 meters is written to one significant digit the next part of the question was to

calculate the speed of the cannonball 2500 milliseconds after launch

2500 milliseconds is 2.5 seconds and so a little later on we'll say that the ball is approximately

here 2.5 seconds later and we need to solve for the velocity or the speed ultimately

and so we draw our vector and we break that vector apart into an x and a y component

now one of the parts of the drawing one of those vectors we've drawn we actually already know

so i want you to really give this some thought which vector which vector do we already

know pause the video now think about it all right i hope you gave it some

thought we know the x component why is this remember we said that along the x direction there

are no forces acting and if there's no forces acting that means the velocity in the x

direction cannot change so whatever velocity we start off with in this case it was 72 kilometers an

hour or 20 meters per second that velocity remains constant regardless

of whatever position the ball is at it's always going to be 20 meters per second

in the x direction and so our goal of course is still to get v2 the final speed

to do this we need to solve for the velocity in the y direction so focusing on the y direction

now we know the time of flight is two and a half seconds

the acceleration is 9.8 meters per second per second remember that the initial velocity in

the y direction what is it well we look at that point and notice again as already mentioned at that

specific point all of the motion is in the x direction it's not moving up or down when it's

launched and so the initial velocity in the y direction is 0

meters per second now i'd like you to use that information pause the video

and solve for the final velocity i hope you tried it here is the final velocity in the y

direction and so substituting that number into our triangle now at this point i

leave it to you to use pythagorean theorem our final velocity works out to 31.6

meters per second or 30 meters per second when written to

significant digits i hope you enjoyed today's video please watch

part two of projectile motion to see another example of projectile motion have a great day bye-bye