2D Motion Projectile Motion

Transcription

1 2D Motion Projectile Motion Lana heridan De Anza College Oct 3, 2017

2 Last time vectors vector operations

3 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

4 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

5 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

6 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

7 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

8 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

9 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

10 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

11 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

12 Overview 2 dimensional motion projectile motion height of a projectile

13 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.

14 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

15 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

16 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.)

17 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

18 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 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

19 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 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

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

21 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 at2 r = 1 2 (v i + v f )t

22 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: This one is a scalar equation: r = v i t at2 r = 1 2 (v i + v f )t v 2 f = v 2 i (Why is it a scalar equation?) + 2a r

23 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?

24 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?

25 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?

26 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?

27 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

28 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

29 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

30 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

31 Height of a Projectile tely om- How can we find the maximum height that a projectile reaches? Find the height when v y = 0. - 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

32 Height of a Projectile tely om- How can we find the maximum height that a projectile reaches? Find the height when v y = 0. - he re c y vi v y 0 v 2 f,y = v 2 i,y 2g y 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

33 Height of a Projectile tely om- How can we find the maximum height that a projectile reaches? Find the height when v y = 0. - he re c tile t i 5 oriolf O y vi u i h R v y 0 Figure 4.9 A projectile launched over a flat surface from the origin at t i 5 0 with an initial velocity x v 2 f,y = v 2 i,y 2g y 0 = v 2 y,i 2gh h = v 2 y,i 2g In the diagram, v y,i = v i sin θ. h = v 2 i sin 2 θ 2g

34 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

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

36 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.

37 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.

38 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. 3 Or, notice that just when striking the ground, y = 0.

39 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

40 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 2. v y,f = v y,i + a y t v i sin θ = v i sin θ gt t = 2v i sin θ g t flight = 2v i sin θ g t flight = 2v i sin θ g

41 Time of Flight of a Projectile 3 Or, notice that just when striking the ground, y = 0. y = v y,i t 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?

42 Time of Flight of a Projectile 3 Or, notice that just when striking the ground, y = 0. y = v y,i t 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? 1 2 gt = v i sin θ t flight = 2v i sin θ g

43 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 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) A 15, 30, 45, 60, 75 B 45, 30, 60, 15, 75 C 15, 75, 30, 60, 45 D 75, 60, 45, 30, 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 x (m)

44 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 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) A 15, 30, 45, 60, 75 B 45, 30, 60, 15, 75 C 15, 75, 30, 60, 45 D 75, 60, 45, 30, 15 1 Page 86, erway & Jewett v i 50 m/s Complementary values of the initial angle u i result in the same value of R x (m)

45 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.