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(II) A ball player catches a ball 3.2 s after throwing it verti-cally upward. With what speed did he throw it, and whatheight did it reach?

$13 \mathrm{m}$

Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Cornell University

Rutgers, The State University of New Jersey

McMaster University

Lectures

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

04:16

In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

03:45

(II) A ball player catches…

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01:50

(II) A ballplayer catches …

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03:41

For a flourish at the end …

01:44

You throw a ball straight …

02:47

A ball is thrown straight …

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When a dropped steel ball …

05:38

A juggler throws a ball st…

00:49

When a ballplayer throws a…

06:01

A soccer player takes a fr…

01:14

05:26

A ball thrown straight up …

02:41

A ball is thrown upward fr…

02:44

You stand at the top of a …

Okay, so someone throws a ball up and it comes back down, and we know the total time it takes to do this is 3.2 seconds. And what else do we know? We know that the acceleration is gonna equal minus G. And if we're going up in all the way back down, then our delta Y is zero because it's starting at the same place it ends. So we plug into this equation Delta y equals V zero t plus 1/2 A T squared. Don't. The Y zero p zero is what we're looking for and we have minus 1/2 g t squared. Now, if I bring this term over the other side, we get a minus B zero t. But that's gonna cancel with the other minus, and we see the tea's also canceled, so the zero equals 1/2 g. T. Now we can just plug in. This is gonna be 9.8 and 3.2 seconds, and this is approximately equal to 16 meters per second. And for the second part, we want to know the maximum height it reaches. And so one additional piece of information we know is that the it's gonna equal zero this final velocity. Once it reaches its maximum height, it's not gonna be moving. We already know the initial velocity about 16 years for second. We know a equals minus g again and Delta y In this case, it's just why, minus y zero and why zero is zero. So this is gonna be why So if I use this equation, don't know why Right now, this Delta y just becomes a why. So we have zero squared equals B zero squared, minus two g. Why? And then we're looking for Okay, we know V zero. So let me just isolated AA minus V zero squared equals minus two g. Why? So we know why the maximum height it reaches its gonna be V zero squared over to G. So why equals 16 squared over two times 9.8 and this is about 13 meters

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