6. ProjectileRevF.pptx

KINEMATICS

2D MOTIONS

Here in Physics 3, you will

encounter three types of 2D

motions

2D MOTIONS

• Projectile Motion

COMPONENTS OF PROJECTILE

MOTION

Projectile motion is just a

combination of HORIZONTAL

PROJECTILE MOTION

•

By combining

the horizontal

and vertical

motion, we get

projectile

motion

•

The path of the

projectile is

PROJECTILE MOTION

 is any body that is given an

Trajectory -

Projectile-

Range-

ASSUMPTIONS:



the free-fall acceleration

g

is

constant over the range of motion

and is directed downward

HORIZONTAL MOTION with CONSTANT VELOCITY VERTICAL MOTION with CONSTANT ACCELERATION

PROJECTILE MOTION

– horizontal motion

 (x-component)

– vertical motion

At any time the distance from the

origin is given by:

The projectile’s speed at any time is:

PROJECTILE MOTION

2

2

y

x

r





2

2

y

x

v

v

Components of

Initial Velocity



The initial velocity, v

o

, of

launch is a vector.

Again, it has two

components, v

and v

.



v

ox

= v

cos θ

V

oy

= v

sin θ



v

v

Examining closely the components of the

velocity in projectile motion, we can make

the ff.

observations

:

The horizontal component of the velocity is the same all throughout the path of the projectile

Only the vertical component of the velocity of the projectile At the top of the trajectorychanges , the velocity only consist

Horizontal and

Vertical UARM



The Horizontal and Vertical motions are

independent

of each other;

FULL PROJECTILE PATH



A projectile is shot at some upward angle from

the origin



Galileo tells us the horizontal motion is just

steady velocity, the vertical motion is the same

as that of a ball thrown directly upwards.



Therefore



Eliminating

t

gives a parabolic curve through

O:

2 1

0x

,

.

x v t y v t







gt









/

/ 2

.

Equations of Projectile

Motion

1. Trajectory Equation:

2. The Equation for

Time of Flight (T):

EXAMPLE 1

A ball is thrown at 20 m/s at

an angle of 55 degrees above

the horizontal. The ball

 Solution:

 Given velocity (Vo) = 20 m/s, launch angle of 55°,

and height the ball was thrown (1.8m), we can find the height of the ball at a distance of 10m using Equation (1) below

 From which

 y(x) = (tan 55°) 10m – (9.8/2*20² cos²o)10²

Robin fires an arrow that

leaves a bow at 20 m/s and

hits a target 30 m away.

What is the angle that he

should point the arrow to

hit the target?

 Solution:



Knowing velocity (Vo) = 20m/s and

range (R) = 30 m, then we can use

Equation (3) below to find the angle

 From which

sin2



o = R *g /Vo² = 30 * 9.8/20²

= 0.735

2



o = sin -¹ 0.735 = 47°



o = 23.5



Equations of

Projectile Motion

4. Maximum Height

(y

):

MAXIMUM RANGE

A

Taking the muzzle velocity as fixed, how do we find the angle of firing to

maximize the range?

Now

 And maximum range

at 45°

http://www.edupics.com/human-cannon-t10746.jpg





0x 0

cos ,

sin

v



v



v



v











2

0 0 0

2 0

2

/

2

/

sin cos

/

sin 2

x y

R

v v

g

v

g

v

g













/

EXAMPLE 3

A projectile is fired horizontally from the

roof of the building 30.0 m high at an

initial velocity of 20.0 m/s. Find (a) the

total time the projectile is in the air,

(b)where the projectile will hit the ground

and (c) the velocity of the projectile as it

hits the ground.

Solution:

a) t = (√2Y/g) = (√2 (30)/9.8) = 2.47 s

b) X = Vxo *t = 20 * 2.47 = 49.4 m

NOTES

•

The maximum range a projectile can

have when it is launched or projected

is at the

angle of projection of 45

•

If an angle,



, result to range R,

another angle,



will yield the same R

given that they

are complements of

each other

NOTES

As can also be observed, each projectile reach different

altitudes and different horizontal ranges.

Without air resistance, the vertical motion of a projectile is governed by the same

principle as in free fall. Thus,

- the speed lost while

going up WILL EQUAL the

speed gained while going down.

- the time going up WILL ALSO EQUAL the time

going down

Without air resistance, the vertical motion of a projectile is governed by the same

principle as in free fall. Thus,

- the speed lost while

going up WILL EQUAL the

speed gained while going down.

- the time going up WILL ALSO EQUAL the time

going down

In the presence

of air

resistance,

however, we

expect the

trajectory of a

high-speed

projectile to fall

short of a

parabolic path

In the presence

of air

resistance,

however, we

expect the

trajectory of a

high-speed

projectile to fall

short of a

parabolic path

EXAMPLE 5

 Suppose a large rock is ejected from a volcano with a speed of

Solution:

The time a projectile is in the air is governed by its vertical motion alone. While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting

altitude.

We can find the time for this by using Eq 2-11b

Y = Yo + Voy*t −1/2g*t²

If we take the initial position Yo to be zero, then the final position is

Y = −20.0 m.

Initial velocity is Voy = Vo sin θo = ( 25.0 m/s )(sin 35.0º) = 14.3 m/s .

So

−20.0 m = (14.3 m/s) t − ½ (9.8) m/s² * t².

Rearranging terms gives a quadratic equation in terms of t :

4.90 m/s t² − (14.3 m/s)t − (20.0 m) = 0.

where the constants are a = 4.90 , b = – 14.3 , and c

= – 20.0.

Its solutions are given by the quadratic formula:

t = (−b ± √(b² − 4ac ))/2a

This equation yields two solutions: t = 3.96 and t = – 1.03 . The negative value of time implies an event before the start of motion, and so we discard it.

Thus,