Loading in 5 sec....

Introduction to Projectile MotionPowerPoint Presentation

Introduction to Projectile Motion

- 47 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Introduction to Projectile Motion' - marcena-terence

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Introduction to Projectile Motion

Introduction toProjectile Motion

Mr.Chin-Sung Lin

Introduction to Projectile Motion

- What is Projectile Motion?

- Trajectory of a Projectile

- Calculation of Projectile Motion

Introduction to Projectile Motion

- What is Projectile Motion?

- Trajectory of a Projectile

- Calculation of Projectile Motion

Features of Projectile Motion?

Thrown into the Air

2-D Motion

Parabolic Path

Affected by Gravity

Determined by Initial Velocity

Definition: Projectile Motion

Projectile motion refers to the 2-D motion of an object that is given an initial velocity and projected into the air at an angle.

The only force acting upon the object is gravity. It follows a parabolic path determined by the effect of the initial velocity and gravitational acceleration.

Definition: Projectile Motion

Projectile motion refers to the 2-D motion of an object that is given an initial velocity and projected into the air at an angle.

The only force acting upon the object is gravity. It follows a parabolic path determined by the effect of the initial velocity and gravitational acceleration.

Introduction to Projectile Motion

- What is Projectile Motion?

- Trajectory of a Projectile

- Calculation of Projectile Motion

x

x

x

- Velocity is changing and the motion is accelerated
- The horizontal component of velocity (vx) is constant
- Acceleration from the vertical component of velocity (vy)
- Acceleration due to gravity is constant, and downward
- a = - g = - 9.81m/s2

g = 9.81m/s2

x

- The horizontal and vertical motions are independent of each other
- Both motions share the same time (t)
- The horizontal velocity ....vx = v0
- The horizontal distance .... dx = vx t
- The vertical velocity ........ vy = - g t
- The vertical distance ........ dy = 1/2 gt2

g = 9.81m/s2

x

Trajectory (Path) of a Projectile

- The path of a projectile is the result of the simultaneous effect of the H & V components of its motion

- H component constant velocity motion

- V component accelerated downward motion

- H & V motions are independent

- H & V motions share the same time t

- The projectile flight time t is determined by the V component of its motion

Trajectory (Path) of a Projectile

- H velocity is constant vx = v0

- V velocity is changing vy = - g t

- H range: dx = v0 t

- V distance: dy = 1/2 gt2

- What is Projectile Motion?

- Trajectory of a Projectile

- Calculation of Projectile Motion

v0

d

g

t

R

Calculation of Projectile Motion- Example: A projectile was fired with initial velocity v0horizontally from a cliff d meters above the ground. Calculate the horizontal range R of the projectile.

v0

d

g

t

R

Strategies of Solving Projectile Problems- H & V motions can be calculated independently
- H & V kinematics equations share the same variable t

v0

d

g

t

R

Strategies of Solving Projectile ProblemsH motion: dx = vx t R = v0 t

V motion: dy = d = 1/2 gt2 t =sqrt(2d/g)

So, R = v0 t = v0 * sqrt(2d/g)

Numerical Example of Projectile Motion

H motion: dx = vx t R = v0 t = 10 t

V motion: dy = d = 1/2 gt2 t =sqrt(2 *19.62/9.81) = 2 s

So, R = v0 t = v0 * sqrt(2d/g) = 10 * 2 = 20 m

V0 = 10 m/s

g = 9.81 m/s2

19.62 m

t

R

V0 = 10 m/s

g = 9.81 m/s2

d

t

20 m

Exercise 1: Projectile ProblemA projectile was fired with initial velocity 10 m/s horizontally from a cliff. If the horizontal range of the projectile is 20 m, calculate the height d of the cliff.

V0 = 10 m/s

g = 9.81 m/s2

d

t

20 m

Exercise 1: Projectile ProblemH motion: dx = vx t 20 = v0 t = 10 t t = 2 s

V motion: dy = d = 1/2 gt2 = 1/2 (9.81)22 = 19.62 m

So, d = 19.62 m

V0

g = 9.81 m/s2

19.62 m

t

20 m

Exercise 2: Projectile ProblemA projectile was fired horizontally from a cliff 19.62 m above the ground. If the horizontal range of the projectile is 20 m, calculate the initial velocity v0 of the projectile.

V0

g = 9.81 m/s2

19.62 m

t

20 m

Exercise 2: Projectile ProblemH motion: dx = vx t 20 = v0 t

V motion: dy = d = 1/2 gt2 t =sqrt(2 *19.62/9.81) = 2 s

So, 20 = v0 t = 2 v0 v0 = 20/2 = 10 m/s

Summary of Projectile Motion

- What is Projectile Motion?

- Trajectory of a Projectile

- Calculation of Projectile Motion

Projectile Motion with Angles

Mr.Chin-Sung Lin

Example: Projectile Problem – H & V

A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the horizontal and vertical components of the initial velocity?

g = 9.81 m/s2

20 m/s

vy

60o

vx

Example: Projectile Problem – At the Top

A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the velocity of the projectile at the top of its trajectory?

v

g = 9.81 m/s2

20 m/s

vy

t

60o

vx

R

Example: Projectile Problem – Height

A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the maximum height that the ball can reach?

g = 9.81 m/s2

20 m/s

vy

h

60o

vx

Example: Projectile Problem - Time

A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. How long will the ball travel before hitting the ground?

g = 9.81 m/s2

20 m/s

vy

t

60o

vx

Example: Projectile Problem – H Range

A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. How far will the ball reach horizontally?

g = 9.81 m/s2

20 m/s

vy

60o

vx

R

Example: Projectile Problem – Final V

A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What’s the final velocity of the projectile right before hitting the ground?

g = 9.81 m/s2

20 m/s

vy

60o

vfx

vx

vfy

vf

Example: Projectile Problem – Max R

A projectile was fired from ground with 20 m/s initial velocity. How can the projectile reach the maximum horizontal range? What’s the maximum horizontal range it can reach?

g = 9.81 m/s2

20 m/s

q

R

Download Presentation

Connecting to Server..