Path of a Projectile Repeat Exercise 97 for a projectile with a path given by the rectangular eq

Path of a Projectile Repeat Exercise 97 for a projectile...

Key Concepts

Vertex of a Parabola

The vertex of a parabola is the point where the function reaches its maximum or minimum value. In projectile motion, the vertex represents the maximum height that the projectile attains, and determining its coordinates is essential for analyzing the overall behavior of the trajectory.

Parabolic Trajectory

A parabolic trajectory is the curved path described by a projectile under the influence of gravity. This curve results from the combination of a constant horizontal velocity and a uniformly accelerated vertical motion, yielding a symmetric path where the highest point marks the vertex of the parabola.

Quadratic Functions

Quadratic functions are mathematical expressions of the form ax² + bx + c, and their graphs are parabolas. In the context of projectile motion, these functions model the vertical displacement of the projectile as a function of horizontal displacement, capturing both the ascent and descent of the object.

Projectile Motion

Projectile motion describes the path that an object follows when it is thrown or propelled near the earth's surface and is subject to a constant acceleration due to gravity. This type of motion involves independent horizontal motion with constant velocity and vertical motion with constant acceleration, resulting in a curved trajectory.

Transcript

00:08 So we are given the rectangular equation for the path of a projectile.

00:18 This equation is y equals 6 plus x minus 0 .08 x squared.

00:42 Now, in part a, we're asked to use the result of the previous exercise to find h, b, not, and theta, and define the parametric equations of the path.

00:59 Well, from exercise 98, we know that the rectangular equation must be of the form.

01:07 For a path of a projectile, negative 16, sikin squared theta over v0 squared times x squared plus tangent theta times x plus h.

01:26 And so if we compare this to our given rectangular equation, we see right away that h is equal to 6.

01:33 We know that negative 16 secund squared theta over v0 squared is equal to negative 0 .08 and the tangent theta is equal to 1.

01:59 So we already know what h is.

02:01 To find v .0, we're going to have to know theta, and you can find out theta from the third equation.

02:07 So from the third equation, we have the theta is the inverse tangent of 1, which is just, pi over 4 or 45 degrees.

02:24 And therefore we have v .0 squared.

02:29 This is equal to 16 secant squared of 45 degrees over 0 .08.

02:44 So that v .0, well, this is supposed to be positive, so it's the positive square root.

02:49 This is the positive square root of 16 secan squared of 45 degrees over 0 .0 .0.

02:57 08 and we've plugged this into a calculator.

03:11 The result is 20...

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