Projectile motion (part 2) A derivation of a new motion equation
Projectile motion (part 2)
- In the last video, I dropped myself or a penny from the top
- of a cliff.
- We started off at 0 velocity, obviously because it was
- stationary, and at the bottom, it was 100 meters per second.
- We used that to figure out how high the cliff was, and we
- figured out that the cliff was 500 meters high.
- What I want to do now is let's do that same problem, but
- let's do in a general form, and see if we can figure out a
- general formula for a problem like that.
- Let's say that you have the same thing, and let's say the
- initial velocity-- you're given the initial velocity,
- you're given the final velocity, you're given the
- acceleration, and you want to figure out the distance.
- This is what you're given, and you want to
- figure out the distance.
- Doing it the exact same way we did in that last presentation,
- but now we're now [INAUDIBLE]
- formulas, we know that the change in distance is equal to
- the average velocity times-- we could actually say the
- change in time, but I'll just say it with time, because we
- always assume that we start with time
- equals 0-- times time.
- We know that the average velocity is the final velocity
- plus the initial velocity divided by 2, so that's the
- average velocity-- let me highlight-- this is the same
- thing as this, and then that times time.
- What's the time?
- You could figure out the time by saying, we know how fast
- we're accelerating, and we know the initial and final
- velocity, so we can figure out how long we had to accelerate
- that fast to get that change in velocity.
- Another way of saying that, or probably a simpler way of
- saying that, is change in velocity, which is the same
- thing as the final velocity minus the initial velocity is
- equal to acceleration times time.
- If you want to solve for time, you could say the time-- if I
- just divide both sides of this equation by a-- is equal to vf
- minus vi divided by a.
- We could take that and substitute that into this
- equation, and remember-- this is all change in distance.
- We say change in distance is equal to-- let me write this
- term in yellow-- vf plus vi over 2.
- Let me write this term in green.
- That's times vf minus vi over a.
- Then if we do a little multiplying of expressions on
- the top-- you might have recognized this-- this would
- be vf squared minus vi squared, and then we multiply
- the denominators over 2a.
- So the change in distance is equal to vf squared minus vi
- squared over 2a.
- That's exciting-- let me write that over again.
- The change in distance is equal to vf squared minus vi
- squared divided by 2 times acceleration.
- We could play around with this a little bit, and if we assume
- that we started distance is equal to 0, we could write d
- here, and that might simplify things.
- If we multiply both sides by 2a, we get-- and I'm just
- going to switch this to distance, if we assume that we
- always start at distances equal to 0.
- di, or initial distance, is always at point 0.
- We could right 2ad-- I'm just multiplying both sides by 2a--
- is equal to vf squared minus vi squared, or you could write
- it as vf squared is equal to vi squared plus 2ad.
- I don't know what your physics teacher might show you or
- written in your physics book, but of these variations will
- show up in your physics book.
- The reason why I wanted to show you that previous problem
- first is that I wanted to show you that you could actually
- figure out these problems without having to always
- memorize formulas and resort to the formula.
- With that said, it's probably not bad idea to memorize some
- form of this formula, although you should understand how it
- was derived, and when to apply it.
- Now that you have memorized it, or I showed you that maybe
- you don't have to memorize it, let's use this.
- Let's say I have the same cliff, and it
- has now turned purple.
- It was 500 meters high-- it's a 500 meter high cliff.
- This time, with the penny, instead of just dropping it
- straight down, I'm going to throw it straight up at
- positive 30 meters per second.
- The positive matters, because remember, we said negative is
- down, positive is up-- that's just the convention we use.
- Let's use this formula, or any version of this formula, to
- figure out what our final velocity was when we hit the
- bottom of the ground.
- This is probably the easiest formula to use, because it
- actually solves for final velocity.
- We can say the final velocity vf squared is equal to the
- initial velocity squared-- so what's our initial velocity?
- It's plus 30 meters per second, so it's 30 meters per
- second squared plus 2ad.
- So, 2a is the acceleration of gravity, which is minus 10,
- because it's going down, so it's 2a times minus 10-- I'm
- going to give up the units for a second, just so I don't run
- out of space-- 2 times minus 10, and what's the height?
- What's the change in distance?
- Actually, I should be correct about using change in
- distance, because it matters for this problem.
- In this case, the final distance is equal to minus
- 500, and the initial distance is equal to 0.
- The change in distance is minus 500.
- So what does this get us?
- We get vf squared is equal to 900, and the negatives cancel
- out-- 10 times 500 is 5,000, and 5,000 times 2 is 10,000.
- So vf squared is equal to 10,900.
- So the final velocity is equal to the square root of 10,900.
- What is that?
- Let me bring over my trusty Windows-provided default
- It's 10,900, and the square root.
- It's about 104 meters per second, so my final velocity
- is approximately-- that squiggly equals is
- approximately-- 104 meters per second.
- That's interesting.
- If I just dropped something-- if I just drop it straight
- from the top-- we figured out in the last problem that at
- the end, I'm at 100 meters per second.
- But this time, if I throw it straight up at 30 meters per
- second, when the penny hits the ground, it's actually
- going even faster.
- You might want to think about why that is, and you might
- realize it.
- When I throw it up, the highest point of the penny--
- if I throw it up at 30 meters per second, the highest point
- of the penny is going to be higher than 500 meters-- is
- going to make some positive distance first, and then it's
- going to come down, so it's going to have even more time
- to accelerate.
- I think that makes some intuitive sense to you.
- That's all the time I have now, and in the next
- presentation, maybe I'll use this formula to solve a couple
- of other types of problems.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
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This is great, I finally understand quadratic functions!
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At 2:33, Sal said "single bonds" but meant "covalent bonds."
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