Uniersity of Babylon College of Engineering Mechanical Engineering Dept. Subject : Mathematics III Class : nd First Semester Year : 16-17
VECTOR FUNCTIONS
SECTION 13. Ideal Projectile Motion
Ideal Projectile Motion What do we mean by the word ideal here? Assumptions: The projectile behaes like a particle moing in a ertical coordinate plane. The only force acting on the projectile during its flight is the constant force of graity. The projectile will follow a trajectory that is perfectly parabolic.
Ideal Projectile Motion Position, elocity, acceleration due to graity, and launch angle at t = Initial elocity: sin j cos i sin j cos i sin j Initial position: r i j r = at time t = cosi a gj
o Newton s second law of motion: f f ma dr m dt The force of graity is: f mg Force is in the downward direction j
Newton s second law of motion: the force acting on the projectile is equal to the projectile s mass times its acceleration. For ideal motion, this force is solely the graitational force: m d dt r mgj d dt r gj a gj
Ideal Projectile Motion Position, elocity, and acceleration at a later time t xy, r xi yj a gj R Horizontal range
m Ideal Projectile Motion Newton s second law of motion: the force acting on the projectile is equal to the projectile s mass times its acceleration. For ideal motion, this force is solely the graitational force: d dt r mgj Let s sole this initial alue problem, using initial conditions: and First integration: Second integration: d dt r gj dr r r dt when t dr gt j dt 1 r j r gt t
Preious equations: Ideal Projectile Motion 1 r j r gt t cos i sin j Substitution: r i j 1 r j cos i sin j gt t t 1 cos ti sin t gt j
Ideal Projectile Motion t t 1 gt r cos i sin j This is the ector equation for ideal projectile motion. The angle is the projectile s launch angle (firing angle, angle of eleation), and is the projectile s initial speed. Break into a pair of scalar equations: y sin t gt x cos t These are the parametric equations for ideal projectile motion. Graity constants: m ft g 9.8 3 sec sec 1
Ideal Projectile Motion t t 1 gt r cos i sin j y sin t gt x cos t If the ideal projectile is fired from the point of the origin: cos 1 x x t y y sin t gt x, y 1 instead
Practice Problem A projectile is fired from the origin oer horizontal ground at an initial speed of 5 m/sec and a launch angle of 6 degrees. Where will the projectile be 1 sec later? Parametric equations for this situation, ealuated at this time: x 5cos61 5 1 y 5sin 6 1 9.8 1 384.17 Ten seconds after firing, the projectile is about 384.17 m in the air and 5 m downrange
Height Flight Time Range
Height, Flight Time, and Range The projectile reaches its highest point when its ertical elocity component is zero: dy dt 1 y sin t gt sin gt t sin g Height at this time: sin 1 sin y sin g g g sin g g sin g sin g
Height, Flight Time, and Range To find the total flight time of the projectile, find the time it takes for the ertical position to equal zero: 1 y sin t gt 1 t sin gt sin t, t g
Height, Flight Time, and Range To find the projectile s range R, the distance from the origin to the point of impact on horizontal ground, find the alue of x when t equals the flight time: R cos t g sin sin g x cos t sincos g sin g sin 1 45 Note: The range is largest when or.
Height, Flight Time, and Range For ideal projectile motion when an object is launched from the origin oer a horizontal surface with initial speed and launch angle : Maximum Height: Flight Time: Range: R t y max sin g sin g sin g
Practice Problems Find the maximum height, flight time, and range of a projectile fired from the origin oer horizontal ground at an initial speed of 5 m/sec and a launch angle of 6 degrees. Then graph the path of the projectile. 5sin 6 Max Height: y 9566.37 max 9.8 m 5 sin 6 Flight Time: t 88.37 9.8 sec 5 Range: R sin1,9.485 9.8 m
Practice Problems Find the maximum height, flight time, and range of a projectile fired from the origin oer horizontal ground at an initial speed of 5 m/sec and a launch angle of 6 degrees. Then graph the path of the projectile. Parametric equations for the path of the projectile: x 5cos6 t 5t 1 y 5sin 6 t 9.8 t 5 3 t4.9t Graph window: [, 5] by [, 1]
y Practice Problems To open the 199 Summer Olympics in Barcelona, bronze medalist archer Antonio Rebollo lit the Olympic torch with a flaming arrow. Suppose that Rebollo shot the arrow at a height of 6 ft aboe ground leel 3 yd from the 7-ft-high cauldron, and he wanted the arrow to reach a maximum height exactly 4 ft aboe the center of the cauldron. 6 y max 74 x 9
Practice Problems y (a) Express max in terms of the initial speed and firing angle. y max (b) Use alue of. sin g ymax 74 sin ft and the result from part (a) to find the 6 74 sin 6 3 sin 68 64 sin 435
Practice Problems cos x x cos t (c) Find the alue of. Values when the arrow reaches maximum height: 9 cos sin g 435 9 cos 3 cos 88 435
Practice Problems (d) Find the initial firing angle of the arrow. sin 435 Put these parts together: 88 cos 435 sin cos tan 435 88 435 435 88 68 45 68 tan 1 45 56.55