2D Motion Projectile Motion - PDF Free Download

2D Motion Projectile Motion Lana heridan De Anza College Oct 3, 2017

Last time vectors vector operations 2 dimensional motion

Warm Up: Quick review of Vector Expressions Let a, b, and c be (non-null) vectors. Let l, m, and n be non-zero scalars. Could this possibly be a valid equation? a = b (A) yes (B) no

Warm Up: Quick review of Vector Expressions Let a, b, and c be (non-null) vectors. Let l, m, and n be non-zero scalars. Could this possibly be a valid equation? a = n (A) yes (B) no

Warm Up: Quick review of Vector Expressions Let a, b, and c be (non-null) vectors. Let l, m, and n be non-zero scalars. Could this possibly be a valid equation? a = b + n (A) yes (B) no

Warm Up: Quick review of Vector Expressions Let a, b, and c be (non-null) vectors. Let l, m, and n be non-zero scalars. Could this possibly be a valid equation? a = n b (A) yes (B) no

Overview projectile motion height and range of a projectile trajectory equation for a projectile

Motion in 2 Dimensions The Plan: All the same kinematics equations apply in 2 dimensions. We can use our knowledge of vectors to solve separately the motion in the x and y directions.

such as Dt placement Motion in 2 Dimensions conclude are Equan. That is ends only with onepoint and ecause its Figure 2.2 back and sional suracross the ry compliver, a play- O y ri The displacement of the particle is the vector r. t i rf r t f Path of particle Figure 4.1 r A = particle xi + yj moving in the xy plane is located with the position r vector = r f r rdrawn i from x

Motion in 2 Dimensions apter 4 Motion in Two Dimensions As the end point approaches, t approaches zero and the direction of r approaches that of the green line tangent to the curve at. y Direction of v at r 1 r 2 r 3 As the end point of the path is moved from to to, the respective displacements and corresponding time intervals become smaller and smaller. O x r(t + t) r(t) = dr icle moves average ion of the r. By definis velocity at line tan- As a particle moves from one point to another along some path, its instantaneous velocity vector v changes = limfrom vi at time t i to vf at time t f. Knowing the velocity t 0 t dt at these points allows us to determine the average acceleration of the particle. The average acceleration a avg of a particle is defined as the change in its instantaneous

Velocity in 2 Dimensions Different directions are independent differentiate separately! r = xi + yj v = dr dt = dx dt i + dy dt j v = v x i + v y j (Differentiation is a linear operation.)

hen a particle accelerates. First, the magnitude ) Acceleration may change with in 2time Dimensions as in straight-line (one- ion to m v i to v f. how two the initial O y r i r f v i v v f v i v f or v f v i x v v = v(t + t) v(t) = v f v i v(t + t) v(t) a = lim = dv t 0 t dt

r Kinematic Equations in 2 Dimensions same reason, from Figure 4.5b we see that rf is gen i, vi, or a. Finally, notice that vf and rf are gener y v yf a y t v f at y f a y t 2 1 2 v f = v i + at Figure 4.5 Vector representations and components of (a) the velocity and (b) the position of a particle under constant acceleration in two dimensions. a v yi v xi v i v xf a x t x b v yi t y i

r Kinematic Equations in 2 Dimensions same reason, from Figure 4.5b we see that rf is gen i, vi, or a. Finally, notice that vf and rf are gener y v f = v i + at v yf a y t v f at y f a y t 2 1 2 v f = (v x,i i + v y,i j) + (a x i + a y j)t Figure 4.5 Vector representations and components of (a) the velocity and (b) the position of a particle under constant acceleration in two dimensions. v x i + v y j = (v x,i + a x t)i + (v y,i + a y t)j Equating x-components (i-components): a v yi v xi v i v xf a x t x b v yi t y i v x = v x,i + a x t Equating y-components (j-components): v y = v y,i + a y t

Kinematic Equations in 2 Dimensions The other kinematics equations work basically the same way as v f = v i + at.

Kinematic Equations in 2 Dimensions The other kinematics equations work basically the same way as v f = v i + at. These are also vector equations and the components add up the same way: r = v i t + 1 2 at2 r = 1 2 (v i + v f )t

Projectiles projectile Any object that is thrown. We will use this word specifically to refer to thrown objects that experience a vertical acceleration g. Assumption Air resistance is negligible. Why do we care?

Projectile Velocity 4 v y i y The y component of velocity is zero at the peak of the path. vi u i v xi v y u v v xi v y 0 v The projectile is launched with initial velocity v i. v y g v xi u v v y The x component of velocity remains constant because there is no acceleration in the x direction. v xi u i x v ection 4.2, we stated that two-dimensional motion with constant accele

Projectile Velocity 4 v y i y The y component of velocity is zero at the peak of the path. vi u i v xi v y u v v xi v y 0 The projectile is launched with initial velocity v i. But the y acceleration is not zero! v v y g v xi u v v y The x component of velocity remains constant because there is no acceleration in the x direction. v xi u i v x ection 4.2, we stated that two-dimensional motion with constant accele

Vector Addition can give a Projectile s Trajectory eleraand y ed in a y 5 e two ction O y vit rf 1 t 2 2 g (x,y) x (Eqs. Figure 4.8 The position vector r f of a projectile r = r f 0 = launched v i t + 1 from 2 at2 the origin whose initial velocity

Height of a Projectile tely om- How can we find the maximum height that a projectile reaches? - he re c y vi v y 0 tile t i 5 oriolf O u i h R Figure 4.9 A projectile launched over a flat surface from the origin at t i 5 0 with an initial velocity x

Time of Flight of a Projectile ion are time completely of flight time The t as time the com- from launch to when projectile hits the ground. How can we find the time of flight of a projectile? arabolic path y ty and accelera- ) nowhere (b) the v y 0 at what point are to each other? v tile l case of projectile the origin at t i 5 O i u i rns to the same horifootballs, Assuming and that golf it is over launched a flat surface from the from ground the origin and lands on the Figure 4.9 A projectile launched. ground at the sameat height... t i 5 0 with an initial velocity h R x

Time of Flight of a Projectile Use symmetry! A parabola is always symmetric about a line through it s vertex.

Time of Flight of a Projectile Use symmetry! A parabola is always symmetric about a line through it s vertex. 1 That means the time to go up and return to the ground are the same. Find the time to reach the v y = 0 point, ( t half ) then multiply by two. 2 Or, notice that just before striking the ground, v y,f = v y,i.

Time of Flight of a Projectile 1. Use symmetry! A parabola is always symmetric about a line through it s vertex. 1 That means the time to go up and return to the ground are the same. Find the time to reach the v y = 0 point, ( t half ) then multiply by two. 2 Or, notice that just before striking the ground, v y,f = v y,i. v y,f = v y,i + a y t 0 = v i sin θ gt half t half = v i sin θ g t flight = 2v i sin θ g

Time of Flight of a Projectile 3 Or, notice that just when striking the ground, y = 0. y = v y,i t + 1 2 a y t 2 0 = v i sin θt 1 2 gt2 Now cancel a factor of t. Warning! This will remove one solution to this equation in t. What is it?

such as 75 and 15. Of course, the maximum height and time of flight for one of these values of u i are different from the maximum height and time of flight for the complementary value. Time of Flight of a Projectile Quick Quiz 4.3 1 Rank the launch angles for the five paths in the figure with respect to time of flight from the shortest time of flight respect to time of flight from the shortest time of flight to the longest. to the longest. (Assume the magnitude v i remains the same.) Q uick Quiz 4.3 Rank the launch angles for the five paths in Figure 4.10 with y (m) 150 100 50 75 60 A 15, 30, 45, 60, 75 B 45, 30, 60, 15, 75 C 15, 75, 30, 60, 45 D 75, 60, 45, 30, 15 45 30 1 Page 86, erway & Jewett 15 v i 50 m/s Complementary values of the initial angle u i result in the same value of R. 50 100 150 200 250 x (m)

Range of a Projectile range The distance in the horizontal direction that a projectile covers ion are before completely hitting the ground. time t as the com- How can we find the range of a projectile? arabolic path ty and accelera- ) nowhere (b) the at what point are to each other? y vi v y 0 tile l case of projectile the origin at t i 5 rns to the same hori- O u i h R Figure 4.9 A projectile launched x

Range of a Projectile e s motion are completely ely, with time t as the coms in its parabolic path e velocity and accelerather? (a) nowhere (b) the choices, at what point are parallel to each other? y vi v y 0 O ui h R x Projectile special case of projectile ed from the origin at t i 5 nd returns to the same hori- Figure 4.9 A projectile launched seballs, over a flat surface from the origin There footballs, is no and acceleration golf in the x-direction! (a at t i 5 0 with an initial velocity x = 0) aunched. analyze: the peak point, vi. The maximum height of the projectile x is h, = and vthe horizontal t, which has coordinates x t range is R. At, the peak of the projectile, and the distance trajectory, the particle has coordi- is the ly in terms We just of v i, need u i, and t. g. But tnates (R/2, time h). of flight!

Range of a Projectile ely, with time t as the coms in its parabolic path e velocity and accelerather? (a) nowhere (b) the choices, at what point are parallel to each other? y vi v y 0 Projectile special case of projectile ed from the origin at t i 5 nd returns to the same horiseballs, footballs, and golf aunched. analyze: the peak point, t, which has coordinates projectile, and the distance ly in terms of v i, u i, and g. O ui h R x Figure 4.9 A projectile launched over a flat x surface = vfrom x t the origin at t i 5 0 with an initial ( velocity ) vi. The maximum height 2vi sin of the θ R projectile = v i is cos h, and θ the horizontal range is R. At, the peak gof the trajectory, the particle has coordinates = 2v (R/2, i 2 sin θ cos θ h). R g R = v i 2 sin(2θ) g

Range of a Projectile A long jumper leaves the ground at an angle of 20.0 above the horizontal and at a speed of 11.0 m/s. How far does he jump in the horizontal direction?

Projectile Trajectory uppose we want to know the height of a projectile (relative to its launch point) in terms of its x coordinate. uppose it is launched at an angle θ above the horizontal, with initial velocity v i. For the x-direction: x = v i cos θt t = x v i cos θ

eight h does a Projectile Motion Example: Figure P4.24 #25, page 103 25. A playground is on the flat roof of a city school, 6.00 m above the street below (Fig. P4.25). The vertical wall of the building is h 5 7.00 m high, forming a 1-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of u 5 53.0 above the horizontal at a point d 5 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the horizontal distance from the wall to the point on the roof where the ball lands. b u d h P4.22 1 erway & Jewett Figure P4.25

ummary projectile motion height, range of a projectile trajectory equation for a projectile Collected Homework! due Friday, Oct 6. (Uncollected) Homework erway & Jewett, Ch 4 Work through example 4.5 (ki Jumper) on page 90. Understand it. Ch 4, onward from page 102. Probs: 7, 11, 15, 21, 29, 37, 39 Read Ch 1-4, if you haven t already.