1. Position and Displacement If we want to locate a point in space we can use a position vector that extends from a reference point to the location we are trying to describe. i, j, k are the unit vectors along the x, y, z, directions x, y, z are the components of the vector along those directions (i, j, k) the displacement is just the change in a particle’s position

2. If we have an initial and final positions described by two position vectors, r i and r f So our displacement is defined as: Let’s look at an example:

4. the definitions for velocity and acceleration are basically the same but you have to remember that these are vectors quantities: average velocity: referring to the previous example, if it took the particle 3 seconds to to the displacement, then the average velocity is: similarly, the instantaneous velocity is: GO TO HITT

5. acceleration follows the same rrr-guments: instantaneous acceleration is: the acceleration vector shows the direction of the acceleration of the particle

6. Motion in more than one dimension. http://www.physicsclassroom.com/mmedia/vectors/mzi.html straight line (line of site) path of bullet path of monkey the monkey begins to fall ad the precise moment when the ball leaves the barrel of the gun The ball and the monkey arrive at the point marked by the red dot at the same time Two types of motion occurring !

7. Monkey: Object under constant acceleration in 1 – dimension Bullet: The motion of the bullet is a combination of motion with a constant velocity (along the x-axis) and motion with constant acceleration (along the y-axis) y x

8. x y  vivi v iy opp v ix adj |v i | hyp To start analyzing this problem we must find the x, and y components of the velocity: Let’ start by looking at an example where the launch angle is zero

9. h = 1 m v 0X = 2 m/s  = 0 x = ? now that we have an expression for time, t we can find the range x

10. h = 1 m v 0X = 2 m/s  = 0 x = ? We could also find the final velocity, v f vfvf

11. Let’s work an example that is a launch angle different from 0, but it returns to the same height that it is launched. A pirate ship is 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v 0 = 82 m/s At what angle,  from the horizontal must a ball be fired to hit the ship? We need an expression for the time, t

12. since the projectile will return to the same height that it left from, h = y –y 0 = 0 so our equation becomes: from here we can solve for time, t factor out a t, and it cancels: now substitute this back into:

13. from trigonometry: this is called the Range equation! DANGER WILL ROBINSON: This equation only works if the projectile returns to the same height that it was launched!!!

15. Now let’s look at an example of projectile motion where the projectile lands at a different elevation from it’s launch height. A stone is projected at a cliff of height h with an initial speed of 42 m/s directed at angle  0 = 60 0 above the horizontal. The stone strikes A 5.5 s after launching. Find: h, the speed of the stone just before impacting A, and the maximum height H reached above the ground.

16. v 0 = 42 m/s = 60 0 total time, t = 5.5 s Let’s start by finding the v and y components of the initial velocity, v 0 (Note: this is always a good place to start!!) Now we can proceed with finding the height of the cliff!

17. v 0 = 42 m/s = 60 0 total time, t = 5.5 s v 0y =36.4 m/s v 0x =21 m/s Now let’s try and find the speed of the rock just before impact. Remember that this will be the magnitude of the final velocity vector. And this vector has both an x and y component.

18. v 0 = 42 m/s = 60 0 total time, t = 5.5 s v 0y =36.4 m/s v 0x =21 m/s 52m Because there is no acceleration along the x axis: v 0x =v fx =21 m/s However in the y-direction there is acceleration so we must find the y component of the final velocity: This negative sign means the y-component is downward! vfvf

19. v 0 = 42 m/s = 60 0 total time, t = 5.5 s v 0y =36.4 m/s v 0x =21 m/s 52m v f =27.3 m/s Now let’s find the maximum height, H! To do this you have to know that the instant the stone is at it’s maximum height the y component of the velocity equals zero, v yMAV H = 0. Using this information we can use:

20. During volcano eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. At what initial speed would a bomb have to be ejected, at angle  0 =35 0 to the horizontal, from the vent at A in order to fall at the foot of the volcano at B, at vertical distance h=3.3 km and horizontal distance d=9.4 km ? = 35 0 So we need to do is figure at what the initial speed of the bombs need to be in order to hit point B. = 9.4 km = 3.3 km

21. = 35 0 = 9.4 km = 3.3 km v 0 =?? We have an equation for range, or horizontal distance: We have, x = 9.4 km, we have  0 =35 0 what we need is the time, t, and this is where it get tricky! Let’s start by using:

22. = 35 0 = 9.4 km = 3.3 km v 0 =?? I’m going to bring the –h over and make this a quadratic equation abc What is the solution to a quadratic equation?

23. = 35 0 = 9.4 km = 3.3 km v 0 =?? a = ½ g b= v 0 sin  0 c = h this reduces to

24. = 35 0 = 9.4 km = 3.3 km v 0 =?? now we can substitute this expression for time into our range equation!

25. = 35 0 = 9.4 km = 3.3 km v 0 =?? Using a lot of algebra and tricks this equation becomes: NOTE: I will ask you to show this on the exam!

26. = 35 0 = 9.4 km = 3.3 km v 0 =255 m/s Next, I would like to find the time of flight. We can rearrange this equation to solve for time: t= ?

27. = 35 0 = 9.4 km = 3.3 km v 0 =255 m/s t= 45 s Finally how would air resistance change our initial velocity? We expect the air to provide resistance but no appreciable lift to the rock, so we would need a greater launching speed to reach the same target.