Projectiles. 50 40 o A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course.

Slides:

Advertisements

Similar presentations

7-2 Projectile Motion. Independence of Motion in 2-D Projectile is an object that has been given an intial thrust (ignore air resistance)  Football,

Advertisements

Projectile Motion. What Is It? Two dimensional motion resulting from a vertical acceleration due to gravity and a uniform horizontal velocity.

1 Projectile Motion. 2 Projectile An object that moves through the air only under the influence of gravity after an initial thrust For simplicity, we’ll.

High School by SSL Technologies Physics Ex-32 Projectile motion is the vectorial sum of two independent velocities, a horizontal component and a vertical.

Jeopardy Vectors 1-D Holdover Concepts Calcu- lations Pictures $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 Final Jeopardy.

Projectile Motion Questions.

Lesson Thread I Full Projectile Motion 12/1/20131Intellectual Property of Mr. Gary.

PH 201 Dr. Cecilia Vogel Lecture 4. REVIEW  Constant acceleration  x vs t, v vs t, v vs x  Vectors  notation  magnitude and direction OUTLINE  2-D.

Chapter 4 Two-Dimensional Kinematics

Projectiles The red ball is given a velocity U at an angle  to the horizontal U  The time taken for the ball to move up and down is the same time as.

Projectile Motion Review.

Jeopardy Vector Components Equations Concepts Calcu- lations Pretty Pictures $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 Final Jeopardy.

Introduction to Projectile Motion

Projectile Motion. What Is It? Two dimensional motion resulting from a vertical acceleration due to gravity and a uniform horizontal velocity.

2-Dimensional Kinematics Unit 2 Presentation 2. Projectile Problems  Projectile Motion: The two- dimensional (x and y) motion of an object through space.

Physics Lesson 6 Projectile Motion Eleanor Roosevelt High School Mr. Chin-Sung Lin.

A soccer ball is kicked into the air. You may safely assume that the air resistance is negligible. The initial velocity of the ball is 40 ms -1 at an angle.

Projectiles Horizontal Projection Horizontally: Vertically: Vertical acceleration g  9.8 To investigate the motion of a projectile, its horizontal and.

Physics Lesson 6 Projectile Motion

10.4 Projectile Motion Fort Pulaski, GA. One early use of calculus was to study projectile motion. In this section we assume ideal projectile motion:

PHYS 20 LESSONS Unit 2: 2-D Kinematics Projectiles Lesson 5: 2-D Projectiles.

Why is it so hard to get rubbish in the bin?

Chapter 3 Kinematics in Two Dimensions. 3.1 – d, v, & a A bullet is fired horizontally. A second bullet is dropped at the same time and at from the same.

Projectile Motion Practice Problems #1:  A ball is fired from a launcher with an initial velocity of 20.0 m·s -1 at an angle of 30.0° to the horizontal.

Motion in Two Dimensions Chapter 7.2 Projectile Motion What is the path of a projectile as it moves through the air? ? . What forces act on projectiles?

CHAPTER 6 MOTION IN 2 DIMENSIONS.

Motion in Two Dimensions Chapter 7.2 Projectile Motion What is the path of a projectile as it moves through the air? Parabolic? Straight up and down?

Motion in Two Dimensions

Physics 111 Projectile Motion 2.0.

Definition of a projectile Examples of projectiles A golf ball hit off the tee A stone kicked off the top of a cliff A tennis ball served horizontally.

Physics.  A projectile is any object that has been launched with no means of controlling its own flight…it is in free-fall motion while at the same time.

Projectile Motion Projectiles The Range Equation.

5.6 Projectiles Launched at an Angle. No matter the angle at which a projectile is launched, the vertical distance of fall beneath the idealized straight-line.

6.2 General projectile motion

Projectile Motion Introduction Horizontal launch.

Projectiles IB Revision. Gravity does not act sideways gravity makes it accelerate downwards The ball moves with a constant horizontal velocity The ball.

A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its velocity is _______. a. entirely.

Aim: How can we solve angular Projectile problems?

Physics Support Materials Higher Mechanics and Properties of Matter

(Constant acceleration)

Motion in Two Dimensions EQ: What is a projectile?

Physics Lesson 6 Projectile Motion

Projectile Motion Properties

How far from the base of the cliff does he land?

Two Dimensional Kinematics

Projectile Motion Modelling assumptions

Final vertical velocity?

A ball is rolling along a flat, level desk. The speed of the ball is 0

Projectile Motion Introduction Horizontal launch.

Projectile Review.

Projectile Motion 2 Launch Angles (symmetrical trajectories)

Projectile Motion AP Physics C.

Projectile Motion.

Find the velocity of a particle with the given position function

The height of the building

Projectile motion Projectile Motion Subject to Gravity Assumptions:

Physics Support Materials Higher Mechanics and Properties of Matter

Projectile Motion Discussion Questions

Motion in Two Dimensions EQ: What is a projectile?

Recall: A horizontal projectile

A projectile launched at an angle

Chapter 3 Jeopardy Review

Projectile Motion Seo Physics.

Topic 9.2 Space Projectile Motion.

Kinematics in Two Dimensions

Projectile motion.

Motion in Two Dimensions

Presentation transcript:

Projectiles

50 40 o A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course.

50 40 o Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2. What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at (3) (4) Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o Find the greatest height reached y = H At max. height H, v y = 0Use equation (4) (1) (2) (3) (4)

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o Find the greatest height reached y = H At max. height H, v y = 0 Find the greatest height reached (1) (2) (3) (4)

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o y = H At max. height H, v y = 0 Find the greatest height reached (1) (2) (3) (4)

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o y = H At max. height H, v y = 0 Find the greatest height reached (1) (2) (3) (4)

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o y = H At max. height H, v y = 0 Now use equation (2) Find the greatest height reached (1) (2) (3) (4)

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o Find time taken to reach the ground y = H In this particular case, using the symmetrical properties of the parabola, t = 2  3.21 = 6.42 secs (1) (2) (3) (4)

A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course o Find the horizontal distance from the start to the landing point y = H 6.42 secs R = x The horizontal distance, or range is R = x. Use (1) (1) (2) (3) (4)

A ball is thrown from the top of a cliff 52m above sea level. With the x axis taken horizontally, the y axis taken vertically upwards and the origin at the point of projection, the velocity of projection is given by Find the speed and angle of elevation of the ball at the start x y (1) (2) (3) (4) Use (3) and (4) with t = 0 Speed Speed = 25ms -1 If the angle of elevation is , then

A ball is thrown from the top of a cliff 52m above sea level. With the x axis taken horizontally, the y axis taken vertically upwards and the origin at the point of projection, the velocity of projection is given by Find the time of the flight x y (1) (2) (3) (4) Find t when y = -52 by substituting in (2)  t = 4 seconds

A ball is thrown from the top of a cliff 52m above sea level. With the x axis taken horizontally, the y axis taken vertically upwards and the origin at the point of projection, the velocity of projection is given by Find the distance from the foot of the cliff to the point where the ball lands x y (1) (2) (3) (4) We have already found that the duration of the fight was 4 seconds, so substitute t = 4 into (1) x = 24(4) = 96 m

A ball is thrown from the top of a cliff 52m above sea level. With the x axis taken horizontally, the y axis taken vertically upwards and the origin at the point of projection, the velocity of projection is given by Find the velocity when the ball lands and hence the speed and angle of entry x y (1) (2) (3) (4) Substitute t = 4 into (3) and (4) Speed Speed = 40.8 ms -1 If the angle of elevation is , then