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Scrambled eggs pipe down or downpipe see eye to eye up to no good.

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Presentation on theme: "Scrambled eggs pipe down or downpipe see eye to eye up to no good."— Presentation transcript:

1 scrambled eggs pipe down or downpipe see eye to eye up to no good

2 3.2 Equations of Kinematics in Two Dimensions

3 3.2 Equations of Kinematics in Two Dimensions – the horizontal component

4 3.2 Equations of Kinematics in Two Dimensions – the vertical component

5 3.2 Equations of Kinematics in Two Dimensions
The x part of the motion occurs exactly as it would if the y part did not occur at all, and vice versa.

6 Projectile Motion Chapter 3 Section 3
Throw something across the room. Ask what happens to it. Ask why?

7 What is a projectile? When a long jumper approaches his jump, he runs along a straight line which can be called the x-axis. When he jumps, as shown in the diagram below, his velocity has both horizontal and vertical components

8 What is a projectile? Projectiles are objects in free fall that have an initial horizontal velocity. Free fall implies that the only force acting on the object is gravity. Initial horizontal velocity results in motion in two dimensions. Examples of projectiles include An arrow flying towards a target A baseball thrown in the air at an angle other than 90o, unless they are thrown straight up. Elucidate above by asking what projectile motion is similar to that we have studied. (1) Ask what force affects bullets after it leaves gun. (2) How is it different from free-fall.

9 Projectile motion Projectiles follow parabolic trajectories.
Trajectory is the path of a flying object. The horizontal and vertical motions are independent. We can write separate equations of motion for each direction. Horizontal motion on the x-axis and the vertical motion on the y-axis Pre-AP: We will prove this shape in class. Do demonstration with ballistic pendulum. Elucidate that vertical motion is the same, regardless of horizontal. Do living projectile motion.

10 What is a projectile? If a ball is thrown horizontally, (yellow) and another ball is dropped (red)-----they will both HIT THE GROUND AT THE SAME TIME!!!!!!!! We shall do a demo of this

11 Solving Projectile Motion Problems
You need to separate the motion down into 2 parts! The first part is the horizontal movement The second part is the vertical movement Keep the two sets of information separate using a table when solving such problems

12 Horizontal motion equations and constants

13 Vertical motion equations and constants

14 Calculating time of fall
If a projectile is launched horizontally, this tells us that the object has an initial horizontal velocity (ie: in the x-axis) For such projectiles, the initial vertical velocity is zero (ie. in the y-axis) (voy=0)

15 Example 1 People in movies often jump from buildings into pools. If a person jumps from the 10th floor (30.0 m) to a pool that is 5 m away from the building, with what initial horizontal velocity must the person jump?

16 Step 1: Write down Givens and unknowns.
You will have to separate your y-components from your x-components using a table X- Component Y - Component Dx = 5.0 m (from building to pool) Dy = m (negative due to downward direction of fall) vox = ? voy = 0 (horizontal projectile) ax = 0 ay = m/s2

17 Step 2: Write down equations that might solve it.
We can use the y equation to solve for the time of the jump. We then use this calculated time in the x equation to solve for the initial velocity. These are the two common equations you will use in this section!!!

18 Step 3: Carry out the solution. Time for the jump
Using the y equation to solve for the time of the jump.

19 Step 4: Carry out the solution
Step 4: Carry out the solution. Using our time to calculate the initial horizontal velocity Using the x equation to solve for the initial horizontal velocity.

20 Practice Problem 1 An autographed baseball rolls off of a 0.70m high desk. And strikes the floor 0.25m away from the desk. How fast did it roll off the table?

21 Practice Problem # 1 Step 1: Write down what’s Given and your unknowns
Practice Problem # 1 Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table) X- Component Y - Component Dx = 0.25 m (from base of desk) Dy = -0.7 m (negative due to downward direction of fall) vox = ? voy = 0 (horizontal projectile) ax = 0 ay = m/s2

22 Step 2: Write down equations that might solve it.
We can use the y equation to solve for the time of the jump. We then use the time in the x equation to solve for the initial velocity.

23 Step 3: Carry out the solution. Time for the jump
Using the y equation to solve for the time of the jump.

24 Step 3: Carry out the solution
Step 3: Carry out the solution. Using our time to calculate the initial horizontal velocity Using the x equation to solve for the initial velocity.

25 Let us review the concepts for Horizontally launched projectiles
Acceleration due to gravity does not affect horizontal motion 𝑎 𝑥 =0 𝑚/ 𝑠 2 Initial horizontal velocity = final horizontal velocity 𝑣 𝑜𝑥 = 𝑣 𝑥 Initial vertical velocity = 0m/s 𝑣 𝑜𝑦 =𝑚/𝑠 𝑎 𝑦 =−𝑔

26 Practice Problem # 2 - A Falling Care Package
3.3 Projectile Motion Practice Problem # 2 - A Falling Care Package The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. Determine the time required for the Care package to hit the ground. What are the magnitude and direction of the final velocity of the Care package?

27 Practice Problem # 1 Step 1: Write down what’s Given and your unknowns
Practice Problem # 1 Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table) X- Component Y - Component Dx = Dy = m (negative due to downward direction of fall) vox = 115 m/s voy = 0 (horizontal projectile) ax = 0 ay = m/s2

28 Step 2: Write down equations that might solve it.
We can use the y equation to solve for the time of the jump. Since 𝑣 𝑜𝑦 =0

29 Step 3: Carry out the solution. Time for the jump
Therefore, using the modified y equation to solve for the time of the jump.

30 Practice Problem # 2 - A Falling Care Package
3.3 Projectile Motion Practice Problem # 2 - A Falling Care Package b) What are the magnitude and direction of the final velocity of the Care package?

31 Practice Problem # 1 Step 1: Write down what’s Given and your unknowns
Practice Problem # 1 Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table) X- Component Y - Component Dx = Dy = m (negative due to downward direction of fall) vox = 115 m/s voy = 0 (horizontal projectile) ay = m/s2 t = 14.6 s

32 Step 2: Write down equations that might solve it.
We then use the time in the y equation to solve for the final vertical velocity. Since 𝑣 𝑜𝑦 =0𝑚/𝑠

33 Step 3: Carry out the solution. Final vertical velocity of the package
3.3 Projectile Motion Step 3: Carry out the solution. Final vertical velocity of the package

34 Step 4: Carry out the solution
Step 4: Carry out the solution. Magnitude and direction of final velocity

35 Class assignment Not H/W
I will not accept late work Class assignment Not H/W

36 Conceptual Example 5 I Shot a Bullet into the Air...
3.3 Projectile Motion Conceptual Example 5 I Shot a Bullet into the Air... Suppose you are driving a convertible with the top down. The car is moving to the right at constant velocity. You point a rifle straight up into the air and fire it. In the absence of air resistance, where would the bullet land – behind you, ahead of you, or in the barrel of the rifle? Explain 

37 Projectiles at an Angle
Section 3. Cont. Projectiles at an Angle

38 Projectiles Launched at an angle
As shown below, projectiles launched at an angle have an initial vertical and horizontal component of velocity.

39 Projectiles Launched at an angle
Suppose the initial velocity vector makes an angle θ with the horizontal The sine and cosine functions can be used to find the horizontal and vertical components of the initial velocity 𝑣 𝑜𝑥 = 𝑣 𝑜 (𝑐𝑜𝑠𝜃) 𝑣 𝑜𝑦 = 𝑣 𝑜 (𝑠𝑖𝑛𝜃)

40 determine the maximum height that the ball attains.
3.3 Projectile Motion Example 6 The Kickoff A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following: determine the maximum height that the ball attains.

41 Step 1: Write down what you know and don’t know (and resolve the vector provided)
=22 m/s =40o X – Component Y - Component ax = 0m/s2 ay = m/s2 Dx = ? Dy = ? Dt = ? Dt = ?

42 Step 2: Decide on a plan or equation to use.
For the horizontal velocity: For the vertical velocity:

43 Step 3: Carry out the plan.
ay vy voy t ? -9.81 m/s2 14.1 m/s

44 The maximum height that the ball attains = 10 m
3.3 Projectile Motion Example 6 The Kickoff A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following: The maximum height that the ball attains = 10 m The time between kickoff and landing

45 Step 1: Write down what you know and don’t know (and resolve the vector provided)
Vi = 22 m/s voy 40o vox X – Component Y - Component ax = 0m/s2 ay = m/s2 Dx = ? Dy = 10.1 m Dt = ? Dt = ?

46 Step 2: Decide on a plan or equation to use.
We can calculate tine for the vertical displacement: What is the equation for horizontal displacement? How did you derive that? What is the equation for vertical displacement? How do we get vix and viy ?

47 Step 3: Carry out the plan.
For the vertical displacement (to maximum height): = 2.87 s What is the equation for horizontal displacement? How did you derive that? What is the equation for vertical displacement? How do we get vix and viy ?

48 The maximum height that the ball attains = 10 m
3.3 Projectile Motion Example 6 The Kickoff A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following: The maximum height that the ball attains = 10 m The time between kickoff and landing = 2.86 s The range ‘R’ of the projectile?

49 Step 1: Write down what you know and don’t know (and resolve the vector provided)
Vi = 22 m/s voy 40o vox X – Component Y - Component ax = 0m/s2 ay = m/s2 Dx = ? Dy = 10 m Dt = 2.86 s Dt = 1.43 s

50 Step 2: Decide on a plan or equation to use.
For the horizontal displacement: What is the equation for horizontal displacement? How did you derive that? What is the equation for vertical displacement? How do we get vix and viy ?

51 Step 3: Carry out the plan.
For the horizontal displacement: = 48.5 m What is the equation for horizontal displacement? How did you derive that? What is the equation for vertical displacement? How do we get vix and viy ?


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