Physics pre-AP

22

1 attvd i

atvv if

advv if 222

Equations of motion:

We assume NO AIR RESISTANCE!

(Welcome to “Physicsland”), therefore…The path of a projectile is a parabola.Horizontal motion is constant velocity.

Vertical motion is in “free-fall”.

Vertical velocity at the top of the path is zero

Time is the same for both horizontal and vertical motions.

The horizontal and vertical motions are independent.

0

constant

x

x

a

v

2m/s 81.9ya

0top

yv

Concepts concerning Projectile Motion:

horizontal or “x” – direction

xxixfx

xixfx

xixx

davv

tavv

tatvd

222

22

1

vertical or “y” – direction

yyiyfy

yiyfy

yiyy

davv

tavv

tatvd

222

22

1

Remember that for projectiles, the horizontal and vertical motions must be separated and analyzed independently. Remember that “ax” is zero and “ay” is acceleration due to gravity “g = 9.81 m/s2”.

0

0

0

A cannon ball is shot horizontally from a cliff.vx

dy

Range, dx

What do we know? For all projectiles…

0

constant

x

x

a

v

2m/s 81.9ya

yx

yinitialy

ttt

vv

0i

Hint: You should always list your known values at the beginning of any problem and assign those values variables.

vx

vx

dy

Range, dx

0

constant

x

x

a

v

2m/s 81.9ya

yx

yi

ttt

v

0

Remember to keep the horizontal and vertical motions separate. Time is the same for both directions. The time to fall (flight time) depends on the height from which the projectile falls.

Knowns:

A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time, t = ? Find range, dx = ? Find final velocity, vf = ?

Vx= 5 m/s

dy=35 m

Range, dx

0

constant

x

x

a

v

2m/s 81.9ya

yx

yi

ttt

v

0

Let’s begin by finding time.

Knowns:Givens:

m35

m/s 5

d

v

y

x

Vx= 5 m/s

dy=35 m

Range, dx

Start with the distance equation…

22

1

22

1

)81.9(035 t

attvd yiy

Now solve for t… sec67.2905.4

35

905.435 2

t

t

Vx= 5 m/s

dy=35 m

Range, dx

Now that we know time, let’s find dx

(the range). Remember that horizontal motion is constant so there is no acceleration. Let’s use the distance formula again. This time in the x – direction…

m 36.13s 67.2m/s 5

22

1

x

xxix

d

tatvd0

Notice: the distance in the horizontaldirection is the horizontal velocity timesthe time.

dx = vx · t

Vx= 5 m/s

dy=35 m

Range, dx

Final velocity is a vector quantity so we must state our answer as a magnitude (speed that the projectile strikes the ground) and direction (angle the projectile strikes the ground). Also remember horizontal velocity is constant, therefore the projectile will never strike the ground exactly at 90°. That means we need to look at the horizontal (x) and vertical (y) components that make up the final velocity.

st 67.2Calculated:

m 36.13xd

Vf

Determining the velocity at impact: there are two components to the final velocity vector. The constant vx and the accelerated vy.

Vx= 5 m/s

dy=35 m

Range, dx

st 67.2

Calculated:

m 36.13xd

vimpact

Vfx = 5 m/s (constant)

Vfy

θθ

So, we have the x-component already due to the fact that horizontal velocity is constant. Before we can find vimpact, we must find the vertical component, vfy.

Calculating vfy:

m/s 2.26

)67.2)(m/s 81.9(0 2

yf

yf

yiyf

v

sv

atvvTo find vfy, remember that vertical motion is in “free-fall” so it is accelerated by gravity from zero to some value just before it hits the ground.

Vx= 5 m/s

dy=35 m

Range, dx

Now we know both the x- and y- components of the final velocity vector. We need to put them together for magnitude and direction of final velocity.

θ

direction 2.79tan

magnitude m/s 7.26

2.265

52.261

2222

f

yfxf

v

vvv

Putting it together to calculate vf:

Final Velocity = 26.7 m/s, 79.2°

Putting everything together:

vi

θ

Since the initial velocity represents motion in both the horizontal (x) and vertical (y) directions at the same time, we cannot use it in any of our equations. Remember, the most important thing about projectiles is that we must treat the horizontal (x) and vertical (y) completely separate from each other. So…we need to separate “vi” into its x- and y- components. We will use the method we used for vectors.

viy

θ

vix

vix

vi

sin

cos

iiy

iix

vv

vv

Now that we have the components

of the initial velocity, we will use only those for calculations.

viy

viθ

A projectile’s path is a parabola, ALWAYS. That means if a projectile is launched and lands at the same height, there will be symmetry. The angle of launch and angle of landing will be equal. The initial velocity and the final velocity will be the same magnitude. Also, that means the components will be the same.

x

fyv

v

fyxf vvv

1

22

tan

viy

vix

vfθ

vfx

vfy

Since the horizontal motion is always at constant velocity…

xxixf vvv

Since the vertical motion is the same as a ball that is thrown straight up or dropped straight down (in free-fall), the y-components are equal and opposite.

yiyf vv

vfθ

vfx

vfy

vi

θStep 1: List known values! Draw and label picture.

Knowns (for all projectiles):

0

constant

x

x

a

v

2m/s 81.9ya

yx

y

ttt

v

0top

Vy top = 0

Viy

Vx

dymax =height

Range, dx

vi

θ

Vx

Vy top = 0

Vy dymax =height

Range, dx

Step 2: Divide initial velocity into horizontal (x) and vertical (y) components.

sin

cos

iiy

iix

vv

vv

Step 3: Find flight time if possible. Use vertical motion.

Keeping the horizontal and vertical motions separate!

vi

θ

Almost every projectile problem can be solved by starting with the displacement equation to solve for time. In this case…

212 0y iy yd v t a t

Now solve for time. Yes, it is a quadratic equation! This will be the time for the entire flight. NOTE: If you want to find maximum height you will only use half the time. If you want to find range, use the total time.

Finding time – Method 1:Since the initial and final vertical positions are both the same, vertical displacement dy = 0.

Vy top = 0

Vy dymax =height

Range, dx

Note: There are three ways to find time for this problem. You may use any of them you wish.

V i= 3

0 m/s

60°Step 1: List known values! Draw and label picture.

Knowns (for all projectiles):

0

constant

x

x

a

v

2m/s 81.9ya

yx

y

ttt

v

0top

Vy top = 0

Viy

Vx

dymax =height

Range, dx

A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.

Given values:

60

m/s 03

iv

V i= 3

0 m/s

60°

Step 2: Divide initial velocity into x- & y- components.

Knowns (for all projectiles):

0

constant

x

x

a

v

2m/s 81.9ya

yx

y

ttt

v

0top

Vy top = 0

Viy

Vx

dymax =height

Range, dx

A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.

Given values:

60

m/s 03

iv

m/s 2660sin30sin

m/s 1560cos30cos

iiy

iix

vv

vv

We can add these to what we know. WE WILL NOT USE THE 30 m/s again in this problem because it is not purely an x- or y- value.

Finding time – Remember that dy = 0 because the projectile is starting and ending at the same level (y-position). So, using the known and given values for this problem and the components we calculated, we can solve for time.

Vy top = 0

v y =

26 m

/s

dymax =height

Range, dx

v i= 3

0 m/s

60°

sec 3.5

905.426

)81.9(260

905.426

2

22

1

22

1

t

tt

tt

tatvd yyiy

A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.

vix =15 m/s

vi

θ

Vy top = 0

Vy

Vx

dymax =height

Range, dx

vf

θvfx

vfy

Now that I know time, I can add it to my list of known, given, and calculated values. To review…

Knowns (for all projectiles):

0

constant

x

x

a

v

2m/s 81.9ya

yx

y

ttt

v

0top

Givens:

Calculated values:

60

m/s 03

iv

sec3.5

m/s 26

m/s 15

t

v

v

yi

xi

A football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.

vi

θ

Vy top = 0

dymax =height

Range, dx

Now that we know time, let’s calculate horizontal distance. Remember that horizontal acceleration is zero.

vf

θv y =

26 m

/sv

y =-26 m

/sA football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.

vx =15 m/s vx =15 m/s

m 579s) 3.5)( 15(

22

1

.d

tatvd

sm

x

xxix

0

vi

θ

Vy top = 0

dymax =height

Range, dx

To find maximum height for this problem, remember that vytop = 0 So:

vf

θv y =

26 m

/sv

y =-26 m

/sA football is kicked from the ground with a speed of 30m/s at and angle of 60°. Find time of flight, range, and maximum height.

vx =15 m/s vx =15 m/s

ym

ys

ms

m

y

yavv

sm

yiyf

5.34

)(2

2

62.19)26(

) 81.9(2) 26(0

2

2

sm2

22

Vertical displacement is not zero. Consider the launch point as the zero height (reference point) If you know the range: use range to find the time to that position.

Δx = vx t

Use this time in the distance equation to find the height at that position.

22

1 )( tgtvy y

viθ

dy

dx

NOTE: The highest point of the projectile DOES NOT occur at the half-way point of the flight. BE CAREFUL!

Projectiles that do not land at the same height:

vi

θ

-dy

dx

Again, if the range is known use the method in the previous slide.If the range is not known, use quadratic formula to find the time to the landing.Then use this time to find the range.NOTE: The highest point of the projectile DOES NOT occur at the half-way point of the flight. BE CAREFUL!

022

1 tatvd yyiy