 Suppose that a ball is dropped from the upper observation deck of a 250 meter tall building, a tower of some kind. And suppose that the distance traveled in meters after t seconds since you let it go is given by the formula s of t equals 4.9 t squared. And without going into all the physics behind this, let me point out that this is the type of distance function given here on planet Earth that things accelerate due to gravity on planet Earth with respect to this formula. So this is perfectly, perfectly good physics question right here. So let's ask ourselves what is the velocity of the ball after five seconds? So as you drop it, it's going to get faster and faster and faster as gravity is pulling it. How fast is it going to be after five seconds? Well, the key thing to remember here is that velocity in the physical sense is the derivative of positions, the derivative of distance in the situation. That is, we want to remember, this is something that you should all know here, that the velocity is equal to distance prime, so s prime with t. That is to say, velocity is the change of distance with respect to time. Velocity is the first derivative of our function f with respect to time. So if we want to compute the velocity after five seconds, we need to compute the velocity function, which we're going to do that by computing a derivative. Now, we're going to do this by the definition of the derivative, which recall is the limit as h approaches 0 of the difference quotient s of t plus h minus s of t divided by h. So we want to compute this thing right here. Well, what do we know about our function? s is just 4.9 t squared. So we're going to get 4.9 times t plus h squared minus 4.9 t squared. This sits above an h as h goes to 0. You'll notice that in the numerator, the s of t plus h has a factor of 4.9. The s of t also has a factor of 4.9. We could factor this 4.9 out from the numerator, and because it's a limit, we could actually factor this out of the whole limit calculation itself. And so we end up with 4.9 times the limit of t plus h squared minus t squared all over h. You see right here as h goes to 0. And so then we can proceed to do this. We can foil this thing out. We get 4.9 times the limit. Well, what's t plus h squared? You're going to get a t squared plus a 2t h plus a h squared minus a t squared over h as h goes to 0. You'll notice that t squared cancels out with the negative t squared. Those things that didn't cancel out have a factor of h in them. So let's factor out that h. So we get h times 2t plus h all over h. Again, as h is going to 0 right here. We have a factor of h in the numerator. We have a factor of h in the denominator. They will cancel out. Now everything left in the numerator, well, there's no problem of h goes to 0. I'm not going to divide by 0 anymore. So setting h equal to 0, that is allowing h to approach 0 now. We see we get 4.9 times 2t plus 0. This will simplify just to be a 4.9 times 2t, which 2 times 4.9 is going to be 9.8t. And this number, 9.8, is actually significant. On planet Earth, the acceleration due to gravity is 9.8 meters per second squared. So that's significant in this formula right here. Once we have the velocity, we then need to figure out what was the velocity after five seconds. So what we're trying to ask ourselves is what is v of 5? So we take 9.8, we times it by 5, for which that's going to give you 49. And what are the units here? Since the distance was measured in meters with respect to time, which is being measured in seconds, our velocity function, v of 5, is going to be measured as 49 meters per second. This is a nice little trick you see when it comes to derivatives. If you take s of t right here, and this is measured in meters, and then the variable here was measured in time, then when you look at the velocity, which is ds over dt, you can see what's the measurement of distance there that was in meters. What's the measurement of time there was in seconds? So we end up with a unit of meters per second. The units of the derivative will look like a ratio, because after all, derivatives are rates, they're ratios, because you're going to see that there are limits of difference quotients so that derivative will inherit that ratio of units, meters per second here. Another follow-up question we could ask about this is how fast is the ball traveling when it hits the ground? So we're going to need our velocity function, v of t equals 9.8t. We need a plug in a t to figure out, to figure out how fast it's going, but we need to figure out the time when it hit the ground. What is this t sub hit? Well, we don't know. Well, if the ball hits the ground, that means it must have traveled the 250 meters. Remember, we had this 250 meter tower we dropped it off of? Once the distance has gone 250 meters, that's when it's going to hit the ground. So we have to solve the equation s of t equals 250, which remember s of t, this was 4.9t squared. That is equal to 250. So we just solve this algebraically. We're going to divide both sides by 4.9, like so. And then we need to take the square root. So we're going to get t squared is equal to 250 over 4.9. If you don't like the fractions and the decimals together, I mean, that's kind of like, e, that feels so icky. You can always move the decimal over by one in the bottom, which means move it over in the top as well. Basically, what you're doing is your times in the top and bottom by 10 to correct the decimal places there. So you're going to get 2,500 over 49. So taking the square root of both sides, we're going to end up with t is equal to, well, the square root of 2,500 is 50. And then the square root of 49 is going to be 7, like so. And so this is going to be 50 over 7 seconds. Or, you know, we could approximate that. That's going to be approximately 7.14 seconds. So this is the time, this is the time when it's going to hit the ground, like so. Now, I should mention that when you solve an equation with square roots, typically there should be a positive and negative square root coming out here. But as we're talking about time here, what would negative time mean? It's like, oh, it bounced off the ground, came to the top of the tower, and then dropped again. So unless this is like a super rubber ball right now, it's fair to assume that there's only one value, and it's going to be the positive time stamp. So we can go with that. So therefore, the velocity, we need to figure out the velocity at time 7.14. I'm just going to keep it as a fraction to avoid any rounding mistakes right here. 50, 57 times that by 9.8, 50 over 7, which at 50 over 7 might already still be in our calculator, times that by 9.8. Again, if you want to play around with decimals here, this is the same thing as 98 times 5 over 7. Basically, 50 is 10 times 5, 10 times 9.8 would give you 98. And so if you don't want to worry about the decimals, you get something like that. 7 actually does go into 98. It goes in there 14 times. Remember 98 was 49 times 2. So this becomes 14 times 5. Or in other words, this is going to be 70, 70, what 70 miles per or 70 meters per second. Excuse me. So that's how fast that ball is going to be when it hits the ground. So you probably don't want to be the person standing at the bottom of that tower when the ball hits the ground.