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An object is thrown upward from the ledge of a building 128 feet high with an initial velocity of $32 \mathrm{ft} / \mathrm{sec},$ with what speed does it hit the ground?

$$96 \mathrm{ft} / \mathrm{sec}$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 2

Applications of Antidifferentiation

Integrals

Baylor University

University of Michigan - Ann Arbor

Idaho State University

Boston College

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

01:43

An object dropped from a h…

02:29

An object launched upward …

01:02

A rock is thrown down from…

02:25

05:04

An object is launched from…

01:54

An object is launched upwa…

04:44

Use $a(t)=-32$ feet per se…

we have an application of a quadratic problem here that has to do with speed and height um when we're like launching an object. So it's important to remember our base equation which is that the um the height of an object um is equal to this negative one half times G force like this acceleration due to gravity times the time squared um Plus the initial velocity times the amount of time has passed plus some initial height. Um And we know that the acceleration due to gravity we kind of Um are able to leave us 32 ft per second squared. So um In this situation we're given that our initial velocity is 64 feet per second and the initial height as 80 ft. So let's do some substitution and solve our quadratic. So h. of T. is equal to negative 1/2 times 32. So negative 16 T squared Plus our initial velocity 64 times t Plus our initial height 80. Now we can factor this by factoring out the negative 16 and that leaves us with t squared minus 40 minus five. And um to figure out when the projectile hit the ground like when this lands we want to know when the height is equal to zero. So we're going to practice a little bit more. We'll get T times another T. And then two numbers that multiply together to equal negative five add together to equal negative four. That's negative five and positive one. So T -5 and T Plus one. To make this equation true T. Would need to equal either five or T. Would have to equal negative one. We're not going back in time, so it can't be a negative number. And that means that T equals five, which means that the object will land on the ground after five seconds. That's problem # 51.

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