https://youtu.be/-OQroFtfXR0
PHYS 1101: Lecture Five, Part Ten
Let’s do an example. Let’s take this problem I have here. It’s a very straightforward free fall case where a ball is dropped from rest from a tower and it strikes the ground 125 meters below. Approximately how many seconds does it take the ball to strike the ground after it’s dropped?The fact that we can ignore air resistance is excellent. It means we can treat this as just simple free fall with these simplifications to our equations that I mentioned above. But let’s follow our same problem solving steps.
Okay, our first step, visualize. What is this object? The start and the end? Object is the ball that we’re watching and it’s after it’s dropped and just before it strikes the ground. This problem is about free fall and so the scope has to include only that part of the motion for which the object is in free fall, meaning it can’t be touching anything. So it’s just after it’s released and just before it hits the ground. That’s key to the scope for start and end.
Let’s draw a quick motion diagram to see if we picture what’s going on. It starts from rest, so my initial velocity’s going to be 0, but I know every second my velocity vector increases by about 20 miles an hour, 9.8 meters per second squared. I go in here and draw my velocity arrows, and so it’s speeding up. I know this corresponds to acceleration in the same direction, and that’s my acceleration, which is going to be -9.8 meters per second squared. It’s minus because I am going to go with the standard axis definition of positive y up. Move that over a bit for you. We have more room.
Now that we’ve sketched our motion diagram, we’ve got a good visual of velocities and acceleration. I’ve already started step two by drawing an axis. I have to pick an origin, though, in order for my position variables to have meaning. Which should we choose? It doesn’t matter. I’m going to choose, I like working with positive numbers, so I’m going to choose the origin at the ground, so the top of my building is going to be up here where the ball is dropped, or the top of my tower. Let me just do a little sketch here. Here’s my tower.
Okay, so I know that this ball, my object, starts at the top. That’s my initial time position and velocity, and I know that it, at the end of the problem, it’s just about to hit the ground, and at that point I’ll have a time, a position and a final velocity. This final velocity is not 0. It eventually will be 0 after it collides with the ground it comes to rest, but our scope is during free fall and that’s just before it hits the ground.
So, to sketch some vectors to remind myself of that, I wouldn’t really sketch a vector here because it starts at rest, but here, I would sketch a big velocity vector to remind myself that I’m focused on just before it hits the ground. Over, I move that over again a little bit for us so I can now go through my variables and think about which ones I have values for, which ones I know.
Let’s think at the start here, I know I start my clock at 0. What’s my initial y-coordinate? I’ve defined my origin to be at the bottom of the tower. Only when the object is here is its y-coordinate going to be 0. At the top what’s its y-coordinate? It’s in the positive direction from the origin up at the height of the tower, 125 meters. This is the y-coordinate 125 meters. What’s its initial velocity? 0. It starts from rest.
At the end of my scope of my problem, do I know the time? No. In fact, that’s what I need to find. This is what I want. Do I know the final velocity just before it hits the ground? No. I don’t have that. I don’t know it, and I don’t necessarily need to know it.
The next problem solving step, make the list of knowns and the variable to solve for, really done. I added that information right to my drawing. The only variable I didn’t remind ourselves to explicitly address was a. You’ve got to do time, position, velocity, time, position, velocity start and the end, and then what’s a in between? And I already had a written here when I did my motion diagram. So I’ve gone through all seven of those.
Sometimes that’s a good way to approach it. Think of it as a laundry list. Literally go through each one and concentrate on it and say, “Do I have that information? Was it given to me, or is that what I want, I need to solve for?”
Okay, we have our list of knowns. Next step is to pick our equations and try to algebra our way to the solution. What equations, again? They’re our main starting equations that we are customizing to this problem to solve it. I’m going to paste them here just so they’re at a spot where I can see them and kind of work with them. I want to look at the knowns at the same time.
So we want the variable t, let me highlight in yellow, this time, what we want. Only equations one and two have t in it, so right away we’re restricted to one of these. What are the things that we know? Well, we know a because it’s a free fall object. We know the initial velocity, and these are the y variables. We do know the initial y-coordinate and the final. We don’t know the final v.
My eye is drawn to equation two. Every variable we either know or it’s the variable we need to solve for. I would use equation two. It looks straightforward.
But let’s translate that into our y variables just so we don’t get confused. I’m going to kill off 0’s here first. I know my initial velocity is 0, so 0 times t will give me 0. Which of these position coordinates is 0? I define my origin at the bottom of the tower, so y is 0.
My equation now is 0 equals y0 plus one-half at2. I’m focused on the variable t, so I want to reduce my algebra, right, until I end up with an equation I can read. t equals what? Let me first, to try to isolate t, let me go ahead and subtract y0 from both sides. That cancels that. I now have –y0 is equal to one-half at2. I’m going to multiply both sides by 2. That will cancel and move this 2 over. I then need to move over the a, so I’m going to divide both sides by a, and that cancels a from the right hand side.
Now what do I have? Let’s tidy this up. I’m going to move my negative outside, so 2 times –y0 is the same as this, divided by a is equal to t2. I haven’t solved for t yet. I need to undo this square, and I do that by taking the square root.
This gives me then that the square root of -2y0 over a is equal to t. Flipping that around so it reads left to right, t, the quantity I want is equal to -2y0 over a, all of which I have numbers for these variables.
I’m ready to plug things in. t is equal to the square root of -2 times y0. That was the coordinate at the top of the tower plus 125 meters. Plus 125 meters. And then I have to divide by my scaler component, the vector a. That has to include the right sign.
Let me go back and remind you, a for free fall always points down, so it’s got to be -9.8 meters per second squared. -9.8 meters per second squared. The negative there is fortunate because this negative cancels with that, and you need that when you take the square root of a number. The number needs to be positive.
Let’s check units here and see how that works out. This unit of meter in the denominator cancels with that unit of meter, and you might remember here that a fraction in the denominator is the same as flipping it upside down, so this second squared in the denominator of the denominator, comes up to the numerator. So I have t equal to the square root of 2 times 125 divided by 9.8 and the units I have are seconds squared. That looks good to me because the square root of this power of 2 is going to leave me with just seconds and that is the proper units for time.
When I multiply that out I get 5.05 seconds. And thinking about the value, I’ve already checked the unit’s value. Seems about right. 125 meters, that’s about 1.25 football fields high. It’s pretty tall. Probably would take about 5 seconds to fall.