At time t=0, the baseball player releases a ball with the initial conditions shown in the figure

At time t=0, the baseball player releases a ball with the...

Key Concepts

Time Derivatives in Polar Coordinates

Time derivatives in polar coordinates include quantities like the radial velocity (r?) and acceleration (r?), as well as the angular velocity (??) and angular acceleration (??). These derivatives provide insights into how the distance from the origin and the angular position change over time, which is crucial when analyzing the dynamic behavior of a projectile following a non-linear path.

Conversion Between Cartesian and Polar Coordinate Systems

This concept involves translating positions and vectors between the Cartesian (x-y) coordinate system and the polar (r, ?) coordinate system. The conversion formulas, such as x = r cos? and y = r sin?, are essential when the motion is analyzed in one system and the given initial conditions or required final values are in another.

Polar Coordinates

Polar coordinates represent a point in the plane by a distance from a reference point (r) and an angle relative to a reference direction (?). This system is particularly useful in problems where the motion or geometry has a natural radial symmetry, allowing representation of positions and motions in terms of radial and angular components.

Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject only to gravity and no other forces. This concept involves analyzing the trajectory, time of flight, maximum height, and horizontal range, employing the kinematic equations to understand the object's behavior under gravitational acceleration.

Kinematics Equations

Kinematics equations describe the motion of a body without considering the forces that cause it. They relate displacement, velocity, acceleration, and time. In problems of projectile motion, these equations are used to calculate the position and velocity components at any given time using initial conditions such as initial speed and angle of release.

Transcript

00:01 In this problem, we're considering a baseball player, releasing a ball with the initial conditions that i've tried to write in this figure.

00:09 So we release at times zero with the initial velocity of 100 feet per second, and we're asked to find these 1 -2 -456 responses at t equals 0 .5.

00:21 Okay, a couple of things.

00:24 First, r equals the square root of x squared plus y squared.

00:31 That's the rate value of r at any instant.

00:41 And let's call this equation one in case we need to look at this.

00:46 And then let's get the values of x and y at t equals 0 .5 seconds.

00:54 We're going to have x equals xo plus bx, oops, plus bxt, and that will be 0 plus 100 times the cosine.

01:14 Of 30 degrees times 0 .5 and this will equal 43 .30 feet.

01:29 Okay, let me get my next page.

01:33 Now, the xo are the initial x axis coordinates and the velocity at t equals so seconds.

02:01 So let's do this.

02:03 Y equals y -o plus b -y -t plus one -half a -t squared.

02:12 Y equals we were given 60 plus my initial was 100 times the sign of 30 degrees times 0 .5.

02:29 And then we'll add to that one -half times negative 32.

02:36 0 .17 times 0 .0 .5 squared.

02:44 And this will equal 26 .98 feet.

02:53 Okay.

02:56 We'll plug these into our equation 1, and that will equal the square root of 43 point, i think it was 3 .1.

03:09 Holy cow, i lost it.

03:11 That's okay.

03:18 Oh, i went too many pages.

03:19 That's okay.

03:20 43 .3.

03:25 Got it, 43 .3 squared plus 26 .98 squared equals 51 .0 feet.

03:37 So there's my first arm.

03:39 The angle theta at this is going to be the tangent of theta were equal 26 .98 divided by 43 .3, and theta were equal 31 .93 degrees.

03:58 So that's my first set.

04:09 So r and theta are 51 feet and 31 .3 degrees.

04:20 Now let's find our next value.

04:27 Let's get over here.

04:31 Now let's find the velocity at t equals 0 .5.

04:38 So we're going to have vx will be equal to vo times the cosine of this.

04:47 This will be 100 times the cosine of 30 degrees.

04:56 And that will equal 86 .6 feet per second.

05:01 And the velocity in the x direction will still be 33 .91 is constant.

05:14 Vx is constant.

05:18 And vy will be 100 times the sign of 30 degrees minus 32 .15 or 17 minus 0 .5.

05:33 And this will equal 33 .91 feet per second.

05:39 So now we can get our v and that will be 80 square root of 86 .6 seconds plus 33 .91 seconds squared...

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