Physics Lesson 6

Projectile Motion

Eleanor Roosevelt High School

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

What is Projectile Motion?

Features of Projectile Motion?

2-D Motion

Parabolic Path

Affected by Gravity

Thrown into the Air

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

Trajectory (Path) of a Projectile

Trajectory (Path) of a Projectile

v0

x

y

x

y

x

y

x

y

x

y

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

y

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 g t2

g = -9.81m/s2

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

V component accelerated downward motion

H component constant velocity 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

Horizontally Launched Projectile

H velocity is constant vx = v0

V velocity is changing vy = g t

H range: dx = v0 t

V distance: dy = 1/2 g t2

Introduction to Projectile Motion

What is Projectile Motion?

Trajectory of a Projectile

Calculation of Projectile Motion

Calculation of Projectile Motion

Example: A projectile was fired with initial velocity v0 horizontally from a cliff d meters above the ground. Calculate the horizontal range R of the projectile.

g

R

d

v0

t

Strategies of Solving Projectile Problems

H & V motions can be calculated independently H & V kinematics equations share the same

variable t

g

R

d

v0

t

Strategies of Solving Projectile Problems

H motion: dx = vx t R = v0 t

V motion: dy = d = 1/2 g t2 t = sqrt(2d/g)

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

g

R

d

v0

t

Numerical Example of Projectile Motion

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

V motion: dy = d = 1/2 g t2 t = sqrt(2 *19.62/9.81) = 2 s

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

g = 9.81 m/s2

R

19.62 m

V0 = 10 m/s

t

Exercise 1: Projectile Problem

A 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.

g = 9.81 m/s2

20 m

d

V0 = 10 m/s

t

Exercise 1: Projectile Problem

H motion: dx = vx t 20 = v0 t = 10 t t = 2 s

V motion: dy = d = 1/2 g t2 = 1/2 (9.81) 22 = 19.62 m

So, d = 19.62 m

g = 9.81 m/s2

20 m

d

V0 = 10 m/s

t

Exercise 2: Projectile Problem

A 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.

g = 9.81 m/s2

20 m

19.62 m

V0

t

Exercise 2: Projectile Problem

H motion: dx = vx t 20 = v0 t

V motion: dy = d = 1/2 g t2 t = sqrt(2 *19.62/9.81) = 2 s

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

g = 9.81 m/s2

20 m

19.62 m

V0

t

Summary of Projectile Motion

What is Projectile Motion?

Trajectory of a Projectile

Calculation of Projectile Motion

Projectile Motion with Angles

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

60o

vx

vy

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

60o

vx

vy

Vx = V cos θ = 20 m/s cos 60o = 10 m/s

Vy = V sin θ = 20 m/s sin 60o = 17.32 m/s

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?

g = 9.81 m/s2

R

v

t

20 m/s

60o

vy

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?

g = 9.81 m/s2

R

v

t

20 m/s

60o

vy

vx

V = Vx = 10 m/s

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

h

20 m/s

60o

vy

vx

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

h

20 m/s

60o

vy

vx

Vf2 = Vi

2 + 2gd (0 m/s)2 = (17.32 m/s)2

+ 2 (-9.81 m/s2) d

d = 15.29 m

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

t

20 m/s

60o

vy

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

t

20 m/s

60o

vy

vx

Vf = Vi + gt 0 m/s = 17.32 m/s + (-9.81 m/s2) t

t = 1.77 s 1.77 s x 2 = 3.53 s

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

R

20 m/s

60o

vy

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

R

20 m/s

60o

vy

vx

dx = Vx t (R = 10 m/s )(3.53 s) = 35.3 m

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

60o

vy

vx

vfx

vfy vf

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

60o

vy

vx

vfx

vfy vf

Vfx = Vx Vfx = 10 m/s Vfy = -Vy Vfx = -17.32 m/s

Vf = sqrt (Vfx2 + Vfy

2) = 20 m/s

θ = tan-1 (Vfy/Vfx) = -60o

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

R

20 m/s

q

V & H Velocity Vectors of Projectile

15o

30o

45o60o

75o

Launch Angles of Projectile

The End