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alculus 3 Integrals of Vector Functions Question 2 Read the introduction to the subsection "The Vector and Parametric Equations for Ideal Projectile Motion" (p. 774 - 775). Summarize where the Ideal Projectile Motion Equation (Eq (5) on p. 775) comes from. Note how the answer to question 1 comes into play for this. Do not just state the final answer for this but explain where the formula comes from.

alculus 3 Integrals of Vector Functions Question 2 Read the introduction to the subsection "The Vector and Parametric Equations for Ideal Projectile Motion" (p. 774 - 775). Summarize where the Ideal Projectile Motion Equation (Eq (5) on p. 775) comes from. Note how the answer to question 1 comes into play for this. Do not just state the final answer for this but explain where the formula comes from.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Calculus 3 Integrals of Vector Functions 

Question 2 Read the introduction to the subsection "The Vector and Parametric Equations for Ideal Projectile Motion" (p. 774 - 775). Summarize where the Ideal Projectile Motion Equation (Eq (5) on p. 775) comes from. Note how the answer to question 1 comes into play for this. Do not just state the final answer for this but explain where the formula comes from. 

Chapter 13 Vector-Valued Functions and Motion in Space The first integration gives A second integration gives 13.2 Integrals of Vector Functions; Projectile Motion dr dr = -(gt)j + Vo- r = -1/gt²j+Not+Fo- Substituting the values of vo and ro from Equations (3) and (4) gives r = − 1g1²j + (v₁, cos a‹)ti + (v sin a)tj + 0. Collecting terms, we obtain the following. Ideal Projectile Motion Equation r = (1, cos a)ri + (( sin a)t - 2gt² 1 28²²)i. (5) 775 Equation (5) is the vector equation of the path for ideal projectile motion. The angle a is the projectile's launch angle (firing angle, angle of elevation), and up, as we said before, is the projectile's initial speed. The components of r give the parametric equations x = (1, cos a)t and y = (v sin a)t - gr², (6) where x is the distance downrange and y is the height of the projectile at time + > 0.
Transcribed Image Text:Chapter 13 Vector-Valued Functions and Motion in Space The first integration gives A second integration gives 13.2 Integrals of Vector Functions; Projectile Motion dr dr = -(gt)j + Vo- r = -1/gt²j+Not+Fo- Substituting the values of vo and ro from Equations (3) and (4) gives r = − 1g1²j + (v₁, cos a‹)ti + (v sin a)tj + 0. Collecting terms, we obtain the following. Ideal Projectile Motion Equation r = (1, cos a)ri + (( sin a)t - 2gt² 1 28²²)i. (5) 775 Equation (5) is the vector equation of the path for ideal projectile motion. The angle a is the projectile's launch angle (firing angle, angle of elevation), and up, as we said before, is the projectile's initial speed. The components of r give the parametric equations x = (1, cos a)t and y = (v sin a)t - gr², (6) where x is the distance downrange and y is the height of the projectile at time + > 0.
r = 0 at time=0 CE valcos ari a=-gj (a) vol sin aj (x, y) r-xi+yj R a=-g Horizontal range (b) X FIGURE 13.10 (a) Position, velocity, acceleration, and launch angle at 1 = 0. (b) Position, velocity, and acceleration at a later time f. The Vector and Parametric Equations for Ideal Projectile Motion A classic example of integrating vector functions is the derivation of the equations for the motion of a projectile. In physics, projectile motion describes how an object fired at some angle from an initial position, and acted upon by only the force of gravity, moves in a ver- tical coordinate plane. In the classic example, we ignore the effects of any frictional drag on the object, which may vary with its speed and altitude, and also the fact that the force of gravity changes slightly with the projectile's changing height. In addition, we ignore the long-distance effects of Earth turning beneath the projectile, such as in a rocket launch or the firing of a projectile from a cannon. Ignoring these effects gives us a reasonable approximation of the motion in most cases. To derive equations for projectile motion, we assume that the projectile behaves like a particle moving in a vertical coordinate plane and that the only force acting on the projec- tile during its flight is the constant force of gravity, which always points straight down. We assume that the projectile is launched from the origin at time z = 0 into the first quadrant with an initial velocity vo (Figure 13.10). If vo makes an angle a with the horizontal, then Vo = (Vocos a)i + (v sina). If we use the simpler notation up for the initial speed |vo|, then Vo = (u cos a)i + (v sin a)j. The projectile's initial position is ro = Oi + Oj = 0. (4) Newton's second law of motion says that the force acting on the projectile is equal to the projectile's mass m times its acceleration, or m(d'r/dt) if r is the projectile's position vector and t is time. If the force is solely the gravitational force -mgj, then -mgj and -gj. where g is the acceleration due to gravity. We find r as a function of t by solving the following initial value problem. Differential equation: Initial conditions: d'r di² r = ro =-gj and d²r di² dr = Vo di (3) when t = 0
Transcribed Image Text:r = 0 at time=0 CE valcos ari a=-gj (a) vol sin aj (x, y) r-xi+yj R a=-g Horizontal range (b) X FIGURE 13.10 (a) Position, velocity, acceleration, and launch angle at 1 = 0. (b) Position, velocity, and acceleration at a later time f. The Vector and Parametric Equations for Ideal Projectile Motion A classic example of integrating vector functions is the derivation of the equations for the motion of a projectile. In physics, projectile motion describes how an object fired at some angle from an initial position, and acted upon by only the force of gravity, moves in a ver- tical coordinate plane. In the classic example, we ignore the effects of any frictional drag on the object, which may vary with its speed and altitude, and also the fact that the force of gravity changes slightly with the projectile's changing height. In addition, we ignore the long-distance effects of Earth turning beneath the projectile, such as in a rocket launch or the firing of a projectile from a cannon. Ignoring these effects gives us a reasonable approximation of the motion in most cases. To derive equations for projectile motion, we assume that the projectile behaves like a particle moving in a vertical coordinate plane and that the only force acting on the projec- tile during its flight is the constant force of gravity, which always points straight down. We assume that the projectile is launched from the origin at time z = 0 into the first quadrant with an initial velocity vo (Figure 13.10). If vo makes an angle a with the horizontal, then Vo = (Vocos a)i + (v sina). If we use the simpler notation up for the initial speed |vo|, then Vo = (u cos a)i + (v sin a)j. The projectile's initial position is ro = Oi + Oj = 0. (4) Newton's second law of motion says that the force acting on the projectile is equal to the projectile's mass m times its acceleration, or m(d'r/dt) if r is the projectile's position vector and t is time. If the force is solely the gravitational force -mgj, then -mgj and -gj. where g is the acceleration due to gravity. We find r as a function of t by solving the following initial value problem. Differential equation: Initial conditions: d'r di² r = ro =-gj and d²r di² dr = Vo di (3) when t = 0
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