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# Projectile motion (part 1): Using the equations of motion to figure out things about falling objects

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- Welcome back.
- I'm now going to do a bunch of projectile motion problems,
- and this is because I think you learn more just seeing
- someone do it, and thinking out loud,
- than all the formulas.
- I have a strange notion that I might have done more harm than
- good by confusing you with a lot of what I did in the last
- couple of videos, so hopefully I can undo any damage if I
- have done any, or even better-- hopefully, you did
- learn from those, and we'll just add to the learning.
- Let's start with a general problem.
- Let's say that I'm at the top of a cliff, and I jump--
- instead of throwing something, I just jump off the cliff.
- We won't worry about my motion from side to side, but just
- assume that I go straight down.
- We could even think that someone just dropped me off of
- the top of the cliff.
- I know these are getting kind of morbid, but let's just
- assume that nothing bad happens to me.
- Let's say that at the top of the cliff, my initial
- velocity-- velocity initial-- is going to be 0, because I'm
- stationary before the person drops me or before I jump.
- At the bottom of the cliff my velocity is
- 100 meters per second.
- My question is, what is the height of this cliff?
- I think this is a good time to actually introduce the
- direction notion of velocity, to show
- you this scalar quantity.
- Let's assume up is positive, and down is negative.
- My velocity is actually 100 meters per second down-- I
- could have assumed the opposite.
- The final velocity is 100 meters per second down, and
- since we're saying that down is negative, and gravity is
- always pulling you down, we're going to say that our
- acceleration is equal to gravity, which is equal to
- minus 10 meters per second squared.
- I just wrote that ahead of times, because when we're
- dealing with anything of throwing or jumping or
- anything on this planet, we could just use this constant--
- the actual number is 9.81, but I want to be able to do this
- without a calculator, so I'll just stick with minus 10
- meters per second squared.
- It's pulling me down, so that's why the minus is there.
- My question is: I know my initial velocity, I know my
- final velocity, right before I hit the ground or right when I
- hit the ground, what's the distance?
- In this circumstance, what does distance represent?
- Distance would be the height of the cliff, and so how do we
- figure this out?
- What's the only formula that we know for distance, or
- actually the change in distance, but in this case,
- it's the same thing.
- Change in distance is equal to the average velocity.
- When you learned this in middle school, or probably
- even elementary school, you didn't say average velocity,
- because you always assumed velocity was constant-- the
- average and the instantaneous velocity was
- kind of the same thing.
- Now, since the velocity is changing, we're going to say
- the average velocity.
- So, the change in distance is equal to the average velocity
- times time.
- This should be intuitive to you at this point.
- Velocity really is just distance divided by time, or
- actually, change in distance divided by times change in
- time-- or, change in distance divided by
- change in times is velocity.
- Let me actually change this to change in time.
- Since we always assume-- or we normally assume-- that we
- start at distance is equal to 0, and we assume that start at
- time is equal to 0, we can write distance is equal to
- velocity average times time.
- Maybe later on we'll do situations where we're not
- starting at time 0, or distance 0, and in that case,
- we will have to be a little more formal and say change in
- distance is equal to average velocity the change in time.
- This is a formula we know, and let's see what
- we can figure out.
- Can we figure out the average velocity?
- The average velocity is just the average of the initial
- velocity and the final velocity.
- The average velocity is just equal to the average of these
- two numbers: so, minus 100 plus 0 over 2-- and I'm just
- averaging the numbers-- equals minus 50 meters per second.
- We were able to figure that out, so can
- we figure out time?
- We know also that velocity, or let's say the change in
- velocity, is equal to the final velocity minus the
- initial velocity.
- This is nothing fancy-- you don't have to memorize this.
- This hopefully is intuitive to you, that the change is just
- the final velocity minus the initial velocity, and that
- that equals acceleration times time.
- So what's the change in velocity in this situation?
- Final velocity is minus 100 meters per second, and then
- the initial velocity is 0, so the change in velocity is
- equal to minus 100 meters per second.
- I'm kind of jumping in and out of the units, but I think you
- get what I'm doing.
- That equals acceleration times time-- what's the
- acceleration?
- It's minus 10 meters per second squared, because I'm
- going straight down-- minus 10 meters per second squared
- times time.
- This is a pretty straightforward equation.
- Let's divide both sides by the acceleration, by the minus 10
- meters per second squared, and you'll get time is equal to--
- the negatives cancel out, as they should, because negative
- time is difficult, we're assuming positive time, and
- it's good we got a positive time answer-- but the
- negatives cancel out and we get time
- is equal to 10 seconds.
- There we have it: we figured out time, we figured out the
- average velocity, and so now we can figure out the height
- of the cliff.
- The distance is equal to the average velocity minus 50
- meters per second times 10 seconds.
- The distance-- this is going to be an interesting notion to
- you-- the distance it's going to be minus 500 meters.
- This might not make a lot of sense to you-- what does minus
- 500 meters mean?
- This is actually right, because this formula is
- actually the change in distance.
- We said if we did it formally, it would be
- the change in distance.
- So if we have a cliff-- let me change colors with it-- and if
- we assume that we start at this point right here, and
- this distance is equal to 0, then the ground, if this cliff
- is 500 hundred meters high, your final distance-- this is
- the initial distance-- your final disstance df is actually
- going to be at minus 500 hundred meters.
- We could have done it the other way around: we could
- have said this is plus 500 meters, and then this is 0,
- but all that matters is really the change in distance.
- We're saying from the top of the cliff to the ground, the
- change in distance is minus 500 meters.
- And minus, based on our convention, we said minus is
- down, so the change is 500 meters down, and that's height
- of the cliff.
- That's pretty interesting.
- If you go to a 500 meter cliff-- 500 is about 1,500
- feet-- so that's roughly the size of maybe a very tall
- skyscraper, like the World Trade Center
- or the Sears Tower.
- If you jump off of something like that, assuming no air
- resistance, which is a big assumption, or if you were to
- drop a penny-- because a penny has very little air
- resistance-- if you were to drop a penny off of the top of
- Sears Tower or a building like that, at the bottom it will be
- going 100 meters per second.
- That's extremely fast, and that's why you shouldn't be
- doing it, because that is fast enough to kill somebody, and I
- don't want to give you any bad ideas if you're a bad person.
- It's just interesting that physics allows you to solve
- these types of problems.
- In the next presentation, I'm just going to keep doing
- problems, and hopefully you'll realize that everything really
- just boils down to average velocity-- change in velocity
- is acceleration times time, and change in distance is
- equal to change in time times average velocity, which we all
- did just now.
- I'll see you in the next presentation.

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