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Suppose in Exercise $9,$ the ball was thrown vertically upward from a ledge 192 feet high. (a) How high will the ball go? (b) How long does it take for the ball to hit the ground? (c) What is its impact velocity with the ground?

(a) $576 \mathrm{ft}$(b) $12 \sec$(c) -208 ft/sec

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 7

Marginal Functions and Rates of Change

Derivatives

University of Michigan - Ann Arbor

Idaho State University

Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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The next we're gonna look at problem 57 from Chapter two of the Physics 5th. The problem says a ball is thrown straight up and takes 2.25 seconds to reach a height of 36.8 m apart. A What was this initial speed? The part B What is the speed at this height and the part see how much higher will the ball go? So this is the simple setup. We have we have a ball being thrown um for 2.25 seconds and it reaches some .36.8 bus, it might reach some point Rather than 36.8. Um And we will calculate each of these things. So paul a. We need to calculate the initial speed. So we set up our so that's we have sec to 36.8 you is what we want to find. V. Is not important and A -9.8 m A second because we're going in the opposite direction two basement And T. is equal to 2.25 seconds. So we have s. You A and T. So we're gonna use sse youtube plus half a. T squared, just rearranging quickly. U. T. Is going to be equal to s minus half eighties. So we do 36.8 minus a half, multiplied by minus 9.81. Supplied by 2.25 squared Comes out to be 61.6. And therefore are you we just divide by 2.25 to get 27.4 part B. We need to get the speed at this specific height. So this time we don't care about you, we care about e speed at end. So we have our SZ 36.8. You is not important. V. Is what we want to find. Deceleration is minus 9.8. The second squared and time is you 0.25 seconds. This time we're going to use this time since we have SV A and T. We're going to use S equals V. T minus half a T squared. VT much like we did above it, we're going to separate VTR. So V. T. Is equal to S. T. Now plus a half A. T squared compared to or it was minus. So We do v. T is equal to 36.8 plus a half, multiplied by minus 9.8. Multiplied by 2.25 squared. And that comes out to be 12.0 Um and therefore our velocity, we just live by 2.25, We get 5.3 m two part C. Um It asked how much higher will the book go? And for this we have R. S. Is equal to what we want to find. Um you is 27.4 m/s. The is not meters per second A. is -9.8 m/s squared. Um You can also set this up such that let's it's a more direct way of doing it. You don't care about t This is -9.8 and we use B two B zero. And you to be what we just calculated in part B. Um and s is fine. So these are identical methods and they can be used both. Um Ultimately finding the answer. We're just gonna do the slightly longer way around. Um so you squared is equal to -2. So of course with S. U. A. in Vienna, we're going to use the squares. Equity was to S. The U. Squared is you can divide us too, I'm -9.8 Times five, Sorry, Not Times five times s. And therefore S. Is equal to 750, which is the 27.4 spread over 19.6 and therefore S is equal to 38.2 m. Um And therefore, uh We have 38, so 38.2 is the peak of the bulls height. So this is a max height. The question asks, how much further does the ball have to go? So to do that, we just subtract 36.8 from 38.2 to get one point for me to go. The ball has 1.4 m to go. If we were to do this. This other method where we have you squared is equal to minus two, I'm 9.8 -9.8 times S we get s is equal to U squared, which is 28. Wait one over. Um Let's just keep that is 28 28 over 19.6. So we get S. Is 1.4 m, as we found in the other method that you can use either method, whichever works for you. If you prefer to do a bit more methodical, then I recommend this first method. If you will go directly to the answer, you definitely try this man.

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