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A ball is thrown vertically upward, which is the positive direction. A little later it returns to its point of release. The ball is in the air for a total time of 8.0 s. What is its initial velocity? Neglect air resistance.

$+39 m / s$

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

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Lectures

04:34

In physics, kinematics is …

07:57

In mathematics, a position…

05:19

A ball is thrown upward fr…

01:13

03:25

A ball is thrown vertical…

02:00

01:29

An object that is falling …

01:31

02:03

02:26

Physics A ball is thrown v…

04:10

01:05

A ball is thrown verticall…

03:14

You toss a ball straight u…

01:54

$\bullet$ You throw a ball…

A ball is thrown horizonta…

04:09

If a ball is thrown vertic…

04:22

Ball $A$ is thrown vertica…

01:20

04:01

(II) If air resistance is …

04:27

The height (in meters) of …

03:24

09:27

Initial velocity Suppose a…

So for this problem, we have a ball that has thrown vertically upwards a minute is in the air for a total of eight seconds. And after those eight seconds, it returns to its starting position. Now, what this means if an object returns to its starting position, is that that displacement value the X is equal to zero meters because it is no meters away from where it started. Now, as this balls being thrown vertically into the air, we also know its acceleration which is 9.8 meters per second squared towards the ground. Um which, because ground is normally are negative direction. We're going to call negative 9.8 meters per second squared. Now, our question here is asking us to find its initial velocity. So I want to know what the knot is going to be. It means that I'm looking at my equations. I have three numbers I know one I'm looking for and this final velocity, I do not know what it is, and I do not need to solve for it. So when I look over my list of cinematic equations, I'm basically just looking for the equation that does not have that final velocity in it. And we find out that we need to use that last equation. The X equals V not t plus 1/2 A T squared for this problem. So when we plug our numbers in here, we get zero equals, um, eight times that initial velocity minus four point nine times eight squared. Um, when I add, like terms and rearrange this a little bit, we get eight times the initial velocity equals 313.6 and then divide by eight to get our final answer of the ball had to have left the ground at 39.2 meters per second.

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