[MUSIC] This week, we look at two types of motion in two dimension, circular motion and projectiles. Projectiles first, and we'll start with a very old question, which falls more quickly, a heavy object or a light one? Here, an apple and a ring fall at approximately the same rate. For a spanner and a hammer, we can't see a difference. But for a hammer and a feather, a big difference. The difference is due to the force required to push air out of the way. See how this paper bends when I move it through the air, and how slowly it falls? What if we dropped the hammer and feather on the moon where there's no air? Well, a couple of NASA astronauts did that and without air resistance, the hammer and the feather fall at the same rate. >> Watch this. >> A good picture there, I've got. >> Beautiful picture, Dave. >> In my left hand, I have a, a feather. In my right hand, a hammer. And I guess one of the reasons we got here today was because of a gentleman named, Galileo, a long time ago, who made a rather significant discovery about falling objects in gravity fields. And we thought that where would be a better place to confirm his findings than on the moon? And so we thought we'd try it here for you, and the feather happens to be appropriately a falcon feather, for our falcon. And I'll drop the two of them here, and hopefully, they'll hit the ground at the same time. How about that? >> [CROSSTALK]. >> [SOUND] Mr. Galileo was correct in his findings. >> In case you're interested, we provide a resource explaining and quantifying air resistance. But here, we'll concentrate on objects that are heavy enough, small enough, and moving slowly enough that we can neglect air resistance, as we could for the apple, the ring, and the hammer, in our examples. When gravity is the only non-negligible force acting, we say an object is in free fall. I drop a ball from a height, h. I drop a second ball from twice as high, two h. Does the second ball take twice as long to fall? That sounds like a question for you. Here it is in stop frame motion. The balls start from rest, but travel further in each successive frame. They are accelerating downwards. Let's plot vertical position y, and vertical velocity, as functions of time. The slope of velocity is negative. And it seems to be constant. So, let's make that our hypothesis. We write vertical acceleration as ay equals minus g, the acceleration due to gravity. But let's be quantitative. How are the height and the time related for a ball in free fall? Here's the equation we wrote last week for constant acceleration. Let's add the subscript x to show the velocity and the acceleration are in the x direction. Here's the analogous equation for the y direction. We'll use x for horizontal and y for vertical. As you saw, for a dropped object, there's no x velocity. The x0 equals 0. And no x acceleration, ax equals 0. And, if we release it from rest, then the initial y velocity is zero too, vy0 equals 0. Here, we're only interested in the y motion. Let's measure vertical position y from the floor. So the ball hits the floor when y equals 0. Using ay equals minus g, the time when it hits the floor satisfies 0 equals y0 minus 1/2gt squared. So 1/2gt squared equals y0. Rearranging, we see that the square of the falling time is 2y0 over g. So the falling time is proportional to the square root of the initial height. So, assuming constant acceleration, if we double y0, we increase the falling time by root 2 or a factor of about 1.4. If we increase the initial height by 2 squared, I.E., a factor of 4, then the full time increases by a vector of 2. And if we increased it by 3 squared, I.E. by 9, then the full time would increase by a vector of 3. That's interesting. Continuing that series will give us a simple way to measure g and also to see whether it's constant. On this string, there's a mass at position 0, call it mass number zero. Then, mass one is at a height of 15 centimeters, and we'll call that L. Mass two is at 4L, that's 2 squared times L, so it will take twice as long to fall. Mass three is at 9 times L, mass four at 16L. In other words, y equals n squared L where n is the number of the mass. Substitute this for y0, the initial height, then take the n outside the square root. Now we have time to fall is proportional to the number of the ball. They should land equally spaced in time. We can hear they're equally spaced [SOUND] and we can see it in the soundtrack, too. [SOUND] T equals n times square root 2 L over g. So, if I plot T against n, the slope will be square root 2 L over g. We measure the time from the sound track and plot 2 versus n, the number of the masses. So, using L equals 15 centimeters, this gives us an approximate value of g, the magnitude of the acceleration due to gravity alone. The equal spacing of the sounds tells us that the acceleration was pretty constant. Summing up, acceleration due to gravity is downwards which we call the negative y direction. So, if gravity is the only non-negligible force, our equation for motion in the y direction becomes y equals y0 plus vy0 times t minus 1/2gt squared where g equals 9.8 meters per second per second. To get used to using this and related equations, let's do a quiz.