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Motion in 2D (Projectiles!!) Test Friday October 7th

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Presentation on theme: "Motion in 2D (Projectiles!!) Test Friday October 7th"— Presentation transcript:

1 Motion in 2D (Projectiles!!) Test Friday October 7th

2 Review vector addition…

3 Review vector addition…

4 Resultant vectors can be split into vector components
Velocity is a vector quantity, and can be split into a horizontal velocity and a vertical velocity

5 Displacement is also a vector quantity that can be split into components
Displacement starts from an object's initial position and ends at its final position. Route 1 is a direct path represented by a displacement vector. Its tail represents your initial location and its head represents your final destination. Route 2 goes along two roads. You first travel along A, then along B. You end at the same final destination with both routes.

6 Breaking displacement into its vector components
Route 1 has the same displacement as route 2, showing how to add vectors graphically. A and B are perpendicular to each other. The key to breaking a vector into its components is to pick vectors perpendicular to each other that, when added, are equivalent to the original vector.

7 Let’s take a moment to remember some TRIG!
Make sure your calculator is in DEGREES!!! SOH-CAH-TOA

8 Finding the scalar components of a vector from its magnitude and direction
If we know the magnitude of a displacement and the angle (Ө) that the vector makes above or below the positive or negative x-axis, we can determine its scalar components using trig. SOH-CAH-TOA cos 34 = 𝑑 1𝑥 𝑑 1𝑥 =29.85𝑚 sin 34 = 𝑑 1𝑦 𝑑 1𝑦 =20.13𝑚

9 Finding the angle with the x-axis from the vector components
If we know the magnitudes of the vector components or the magnitude of the resultant we can solve for the angles with the x and y axis using inverse trig functions tan 𝜃= 𝜃= tan − 𝜃=36.87°

10 Finding the scalar components of a resultant from its magnitude and direction (using Force as an ex)

11 Turn in your maps from yesterday.
Please have out your notes. We will discuss projectiles today.

12 Projectiles Objects that move vertically and horizontally at the SAME TIME. x and y directions are affected independently vertical (y)—accelerated downward by gravity horizontal (x)—constant velocity Result of two motions on the object give it a parabolic flight path path called a trajectory

13 Examples of Projectiles
Which of these are projectiles?

14 (a) is NOT a projectile (b) IS a projectile
Throw the ball straight up while moving on rollerblades. As long as you do not change your speed or direction while the ball is in flight, it will land back in your hands.

15 Types of Projectile Motion

16

17 Vertical & Horizontal Motion of a Projectile vs. a Dropped Object
Vx = 0 Projectile Vx > 0 Constant velocity Accelerated by gravity Vertical displacements are = for projectiles and dropped objects however horizontal displacements are greater for projectiles. Vertical acceleration is that of gravity (-9.8 m/s2). They both hit the ground at the exact same time but, of course, the projectile is farther away !

18 © 2014 Pearson Education, Inc.
Projectile motion Projectiles are objects launched at an angle relative to a horizontal surface. © 2014 Pearson Education, Inc.

19 Qualitative analysis of projectile motion in the x- and y-axes

20 Qualitative analysis of projectile motion in the y-axis
A ball thrown straight up in the air by a person moving horizontally on rollerblades will land back in the person's hand. Earth exerts a gravitational force on the ball, so its upward speed decreases until it stops at the highest point, and then its downward speed increases until it returns to your hands.

21 Qualitative analysis of projectile motion in the x-axis
A ball thrown straight up in the air by a person moving horizontally on rollerblades will land back in the person's hand. The ball also moves horizontally. No object exerts a horizontal force on the ball, so the ball's horizontal velocity does not change once it is released and is the same as the person's horizontal component of velocity. © 2014 Pearson Education, Inc.

22 Quantitative analysis of projectile motion: Acceleration
The kinematic equations for velocity and constant acceleration can also be used to analyze projectile motion The x-component (in the horizontal direction) of a projectile's acceleration is zero. The y-component (in the vertical direction) of a projectile's acceleration is g (positive or negative based on the direction YOU have chosen)

23 Quantitative analysis of projectile motion: Using our kinematics equations

24 (a) How long will it be in the air?
A ball is thrown horizontally at 12 m/s from a building 30 meters high. Assume the system has no friction. (a) How long will it be in the air? (b) How far from the base of the building will it land? Vox = 12 m/s 30 m vox = 12 m/s Δy = 30 m t = ? voy = 0 m/s

25 A ball is thrown horizontally at 12 m/s from a building 30 meters high
A ball is thrown horizontally at 12 m/s from a building 30 meters high. Assume the system has no friction. (a) How long will it be in the air? (b) How far from the base of the building will it land? ∆xy = v oy t + 1/2 g t 2 Vox = 12 m/s 30 m vox = 12 m/s 30 = 0 + (.5)(9.8)(t2) Δy = 30 m 30 = 4.9t2 t = ? 6.12 = t2 voy = 0 m/s g = 9.8 m/s2 2.47 s = t

26 (a) How long will it be in the air?
A ball is thrown horizontally at 12 m/s from a building 30 meters high. Assume the system has no friction. (a) How long will it be in the air? (b) How far from the base of the building will it land? Vox = 12 m/s 30 m vox = 12 m/s Δy = 30 m t = 2.47 s voy = 0 m/s ? g = 9.8 m/s2 Δx = ?

27 A ball is thrown horizontally at 12 m/s from a building 30 meters high
A ball is thrown horizontally at 12 m/s from a building 30 meters high. Assume the system has no friction. (a) How long will it be in the air? (b) How far from the base of the building will it land? ∆x = vox t Vo = 12 m/s 30 m vox = 12 m/s Δx = (12)(2.47) Δy = 30 m t = 2.47 s Δx = m voy = 0 m/s g = 9.8 m/s2 Δx = ?

28 Example 3.8: Best angle for farthest flight
You want to throw a rock the farthest possible horizontal distance. You keep the initial speed of the rock constant and find that the horizontal distance it travels depends on the angle at which it leaves your hand. What is the angle at which you should throw the rock so that it travels the longest horizontal distance, assuming you throw it with the same initial speed? I'm leaving this for the instructor to solve in class

29 Example 3.9: Shot from a cannon
Stephanie Smith Havens is to be shot from an 8-m-long cannon at 100 km/h. The barrel of the cannon is oriented 45° above the horizontal. She hopes to be launched so that she lands on a net that is 40 m from the end of the cannon barrel and at the same elevation (our assumption). Estimate the speed with which she needs to leave the cannon to make it to the net. This is left for the instructor to solve as an example

30 Ideal vs. Real Projectile Paths
Projectile Path without Air Resistance Projectile Path with Air Resistance


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