Two-dimensional motion Consider the motion of the following...

Two-dimensional motion Consider the motion of the following objects. Assume the $x$ -axis is horizontal, the positive y-axis is vertical, the ground is horizontal, and only the gravitational force acts on the object. a. Find the velocity and position vectors, for $t \geq 0$. b. Graph the trajectory. c. Determine the time of flight and range of the object. d. Determine the maximum height of the object. A bascball has an initial position (in $\mathrm{ft}$ ) of $\left\langle x_{0}, y_{0}\right\rangle=\langle 0,6\rangle$ when it is thrown with an initial velocity of $\left\langle u_{0}, v_{0}\right\rangle=\langle 80,10\rangle \mathrm{ft} / \mathrm{s}$.

Two-dimensional motion Consider the motion of the following objects. Assume the $x$ -axis is horizontal, the positive y-axis is vertical, the ground is horizontal, and only the gravitational force acts on the object.
a. Find the velocity and position vectors, for $t \geq 0$.
b. Graph the trajectory.
c. Determine the time of flight and range of the object.
d. Determine the maximum height of the object.
A bascball has an initial position (in $\mathrm{ft}$ ) of $\left\langle x_{0}, y_{0}\right\rangle=\langle 0,6\rangle$ when it is thrown with an initial velocity of $\left\langle u_{0}, v_{0}\right\rangle=\langle 80,10\rangle \mathrm{ft} / \mathrm{s}$.

Key Concepts

Time of Flight and Range

The time of flight is the total time the projectile remains in the air, determined by its vertical motion under gravity. The range is the horizontal distance traveled during this time, calculated by multiplying the constant horizontal velocity by the time of flight.

Parametric Representation

The parametric representation of motion expresses the horizontal (x) and vertical (y) positions as separate functions of time. This enables the analysis of the trajectory, as the x-coordinate is typically a linear function of time while the y-coordinate is influenced by a quadratic term due to gravity.

Maximum Height

The maximum height is the peak vertical position reached by the projectile. It occurs when the vertical component of the velocity becomes zero, at which point gravity begins to pull the projectile downward. Calculating the maximum height involves analyzing the vertical motion using the kinematic equations.

Vector Decomposition

Vector decomposition involves breaking the initial velocity into its horizontal and vertical components. This is crucial because the horizontal component (which remains constant in the absence of air resistance) and the vertical component (which is affected by gravitational acceleration) govern different aspects of the projectile's motion.

Projectile Motion

Projectile motion refers to the two-dimensional motion of an object that is launched into the air and is only subject to the acceleration due to gravity. This type of motion is characterized by a parabolic trajectory and involves separate motions in the horizontal and vertical directions that occur simultaneously but independently.

Kinematic Equations

Kinematic equations describe the motion of an object under constant acceleration. In the context of projectile motion, these equations are used to determine the position and velocity vectors as functions of time for both horizontal and vertical motions, taking gravitational acceleration into account.

Graphical Trajectory Analysis

Graphing the trajectory involves plotting the vertical position against the horizontal position to visualize the parabolic path of the projectile. This visual representation aids in understanding the motion and determining key characteristics like the maximum height, range, and overall shape of the path.

Transcript

00:01 In this problem of two -dimensional motion, the first thing to note is that we have measurements in feet, meaning that the acceleration of free fall will have components of zero in the x direction and minus 32 feet per second squared.

00:31 When we give an acceleration in meters it's minus 9 .8 meters per second squared when it's in feet it's this all right our job is to find r of t, v of t, graph the trajectory range maximum height and so on we start with a of t being we know that a is derivative of v so v is going to be an integral of a dt.

01:11 What are we given? we are given that the baseball has an initial position, initial position so r of 0 is the initial position of 06 okay what this actually means is that it's from six feet above the ground at an initial position of zero and the initial velocity v of zero is given as 80 in the x direction 10 in the y direction feet per second okay so we start with this v is the end derivative of zero of zero is 0 and the root of minus 32 is minus 32 t the integration is indefinite so we have a constant vector the components we will call p and q v of t is that v of zero is given to be 8010 and on the other hand we have when we plug in zero into this we have this equal 0, minus 32 times 0 is 0 plus pq.

02:49 Equating first components, the first component is 80 is equal to 0 plus p, which is p.

02:57 The second component 10 is equal to 0 plus q, which is q.

03:04 So this is the pq is 80 10.

03:10 Meaning that the velocity function of t is equal to 0 plus p which is 80 minus 32 t plus q which is 10 right so now from from v being the derivative of r, velocity is the derivative of position, it follows that r is going to be the integral of vdt.

03:53 So r of t is obtained by integrating the components of v.

04:03 The anti -derivative of 80 is 80 t, minus 32 t.

04:09 Minus 32 t d t is minus 32 times the integral of t is t squared over two canceling 2 and 32 we have minus 16 t squared so it's minus 16 t squared and the entity derivative of 10 is 10 t plus don't forget the this was the constant vector c1, c2, we will name the components with the nf.

04:52 We are given initial values for r0 to be 06.

05:00 So r of 0, on the one hand, plugging in 0 here, here and here.

05:07 We have 80 times 0 0 0 plus 0 plus ef and as we said we are given it to be 0 6 so equating first components 0 equals 0 plus e which is e and 6 equals 0 plus f which is f so this vector here is 06, meaning that the position of function of t is equal to the first component plus e, which is 0, that is 80t, and the second component minus 16 t squared plus 10 t plus the second component here which is 6, and we have a position function.

06:17 To graph the trajectory, this concludes part a of the problem.

06:23 Alright, let me just move it down a bit...

Please give Ace some feedback