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If a ball is thrown vertically upward with a velocity of $80 \mathrm{ft} / \mathrm{s},$ then its height after $t$ seconds is $s=80 t-16 t^{2}$(a) What is the maximum height reached by the ball?(b) What is the velocity of the ball when it is $96 \mathrm{ft}$ above the ground on its way up? On its way down?

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01:24

Amrita Bhasin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 7

Rates of Change in the Natural and Social Sciences

Derivatives

Differentiation

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

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.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

05:40

if a ball is thrown vertic…

06:19

A ball is thrown verticall…

02:35

Velocity A ball is thrown …

04:10

Physics A ball is thrown v…

01:59

Height of a Ball If a ball…

00:50

in this problem were given the equation for the height of the ball, and we want to find its maximum height. There are two different ways we could do this. The first is sort of the algebra way, so we would recognize that this height equation is that of a parabola, and it would be a problem that opens down. And if we could find its Vertex, we would know the maximum. The other way to do it is to use calculus and to realize that when it's at its maximum, the derivative will be zero. And so that's when the velocity is zero. And my vote is to use calculus for this problem because we're in a calculus class. So we're going to find aware the the time that the velocity is zero. So we need to find the velocity. So we take the derivative of position and we get a T minus 32 T, and then we're going to see when that equals zero. So we set it equal to zero. Keep wanting to say wear it equals zero, but it's when it equal zero. Saw this for tea and we end up with T is 2.5 seconds. So now we want to find the height at 2.5 seconds. So now we go back to our height equation and substitute a 2.5 in there and we get 100 feet. And for Part B, we want to know the velocity when the height is 96 feet and the height is 96 feet in two different times, once when it's on its way up and once when it's on its way down. So the first thing we need to do is find the times that it has a height of 96 feet. So what we can do is take our high equation and set it equal to 96. So we have 96 equals 80 T minus 16 t squared. Let's get all the terms over to one side. So 16 t squared, minus 80 t plus 96 equal zero, and then the whole equation is divisible by 16. So we have t squared minus five. T plus six equals zero, and luckily that's factory Herbal. This is our lucky day. We can factor into T minus three times T minus two and then said each of those factors equal to zero, and we get the two times three seconds and two seconds Notice how those air symmetrical about the time when it was at the top. It was at the top at time, 2.5. So this all makes sense coming together. So now that we know the times, we want the velocities at those times. So the velocity at time to would be a T minus 32 times two, and that is 16 feet per second, and that's when it's on its way up. And then the velocity of time three would be a T minus 32 times three, and that is negative 16 feet per second, and that is on its way down.

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