Projectile motion (part 4) Solving for time when you are given the change in distance, acceleration, and initial velocity
Projectile motion (part 4)
- We'll now use that equation we just derived to go back and
- solve-- or at least address-- that same problem we were
- doing before, so let's write that equation down again.
- Actually, let's write the problem down.
- Lets say I have the cliff again, and so my initial
- distance is 0, but it goes down 500 meters.
- I'm not going to redraw the cliff, because it takes a lot
- of space up on my limited chalkboard.
- We know that the change in distance is equal
- to minus 500 meters.
- I'm still going to use the example where I don't just
- drop the ball, or the penny, or whatever I'm throwing off
- the cliff, but I actually throw it straight up, so it's
- going to go up and slow down from gravity, and then it will
- go to 0 velocity, and start accelerating downwards.
- You could even say decelerating
- in the other direction.
- The initial velocity, vi, is equal to 30 meters per second,
- and of course, we know that the acceleration is equal to
- minus 10 meters per second squared-- it's because
- acceleration gravity is always pulling downwards, or towards
- the center of our planet.
- If we wanted to figure out the final velocity, we could have
- just used the formula, and we did this in the last video--
- vf squared is equal to vi squared plus 2ad.
- Now what I want to do is use the formula that we learned in
- the very last video to figure out-- how long does it take to
- get to the bottom, and to hit the ground?
- Let's use that formula: we derived that the change in
- distance is equal to the initial velocity times time
- plus acceleration time squared over 2, and
- that's initial velocity.
- The change in distance is minus 500, and that's equal to
- the initial velocity-- that's positive, going upwards, 30
- meters per second, 30t.
- I'm not going to write the units right now, because I'll
- run out of space, but you can redo it with the units, and
- see that the units do work out.
- When you square time, you have to square the time units,
- although we're solving for time.
- Then, plus acceleration, and acceleration is minus 10, and
- we're going to divided it by 2, so it's minus 5t squared.
- We have minus 500 is equal to 30t plus minus 5t, and we
- could just say minus 5t squared, and
- get rid of this plus.
- At first, you say, Sal-- there's a t, that's t to the
- first, and t to the second, how do I saw solve this?
- Hopefully, you've taken algebra two or algebra one, in
- some places, and you remember how to solve this.
- Otherwise, you're about to learn the quadratic equation,
- although I recommend you go back, and learn about
- factoring in the quadratic equation, which there are
- videos on that I've put on Youtube.
- I hope you watch those first if you don't remember.
- We can do this-- let's put these two right terms on the
- left hand side, and then we'll use the quadratic equation to
- solve, because I don't think this is easy to factor.
- We'll get 5t squared minus 30t minus 500 is equal to 0-- I
- just took these terms and put them on the left side.
- We could divide both sides by 5, just to simplify things,
- and so we get t squared minus 6t minus 100 is equal to 0.
- I could do that, because 0 divided by 5 is just five, so
- I just cleaned it up a little bit.
- Let's use the quadratic equation, and for those of us
- who need a refresher, I'll write it down.
- The roots of any quadratic-- in this case, it's t we're
- solving for-- t will equal negative b plus or minus the
- square root of b squared minus 4ac over 2a, where a is a
- coefficient on this term, b is the coefficient on this term,
- negative 6, and c is the constant, so minus 100.
- Let's just solve.
- We get t is equal to negative b-- so negative this term.
- This term is negative 6, so if we make it a negative, it
- becomes plus 6.
- It becomes 6 plus or minus the square root of b squared, so
- it's minus 6 squared, 36, minus 4 times a, and the
- coefficient on a is here, and that's just times 1.
- With 4ac, c is a constant term, minus 100, minus 4 times
- 1 times minus 100, and all of that is over 2a-- a is 1
- agains, so all of that is over 2.
- That just equals 6 plus or minus the square root-- this
- is minus 4 times minus 100, and these become pluses, so it
- becomes 36 plus 400.
- So, 6 plus or minus 436 divided by 2.
- This is not a clean number, and if you type into a
- calculator, it's something on the order of about 20.9.
- We can just say approximately 21-- you might want to get the
- exact number, if you're actually doing this on a test,
- or trying to send something to Mars, but for our purposes, I
- think you'll get the point.
- I'll say it approximately now, because we're going to be a
- little off, but just to have clean numbers, this is
- approximately 21-- it's like 20.9.
- We'll say 6 plus or minus-- let me just write
- 20.9-- 20.9 over 2.
- Let me ask you a question: if I do 6 minus
- 20.9, what do I get?
- I get a negative number, and does a
- negative time make sense?
- No, it does not, and that means that somehow in the
- past-- I don't want to get philosophical-- the negative
- time in this context will not make sense.
- Really, we can just stick to the plus, because 6 minus 20
- is negative, so there's only one time that will solve this
- in a meaningful way.
- Time is approximately equal to 6 plus 20.9, so that's 26.9
- over 2, and that equals 13.45 seconds.
- That's interesting.
- I think if you remember way back, maybe four or five
- videos ago, when we first did this problem, we just dropped
- the penny straight from the height.
- Actually, in that problem, I gave you the time-- I said it
- took 10 seconds to hit the ground, and we worked
- backwards to figure out that the cliff was 500 meters high.
- Now, if you're here at the top of a 500 cliff, or building,
- and you drop something that has air resistance-- like a
- penny, that has very air resistance-- it would take 10
- seconds to reach the ground, assuming all of our
- assumptions about gravity.
- But if you were to throw the penny straight up, off the
- edge of the cliff, at 30 meters per second right here,
- it's going to take 13.5-- roughly, 13 and 1/2 seconds--
- to reach the ground.
- It takes a little bit longer, and that should make sense
- because-- I have time to draw a little picture.
- In the first case, I just took the penny, and its motion just
- went straight down.
- In the second case, I took the penny-- it first went up, and
- then it went down.
- It had all the time when it went up, and then it went down
- a longer distance, so it makes sense that this time-- this
- was 10 seconds, while this time was 13.45 five seconds.
- You can kind of say that it took-- well, you actually
- can't say that.
- I don't want to get too involved, but I hopefully this
- make sense to you.
- If you have a smaller number here, you should have gone and
- checked your work, because why would it take less time when I
- throw the object straight up?
- Hopefully, that gave you a little bit more intuition, and
- you really do have in your arsenal now all of the
- equations-- and hopefully, the intuition you need-- to solve
- basic projectile problems.
- I'll now probably do a couple more videos where I just do a
- bunch of problems, just to really drive the points home.
- Then, I'll expand these problems to two
- dimensions and angles.
- Before we get there, you might want to refresh your
- I'll see you soon.
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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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