Projectile Motion

Physics Honors

Good News/Bad News: These are the same formulas we used for linear motion. Do you know them?

221 attvd i atvv if

advv if 222

If the answer is “NO”, then memorize them NOW!

tvvd fi

2

We assume NO AIR RESISTANCE! (Welcome to “Newtonia”), 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.

0

constant

x

x

a

v

downay ,m/s 8.9 2

0top

yv

horizontal or “x” – direction

xxixfx

xixfx

xixx

davv

tavv

tatvd

222

22

1

vertical or “y” – direction

tvvd

davv

tavv

tatvd

yfyiy

yyyiyf

yyiyf

yyiy

2

222

22

1

Remember that for projectiles, the horizontal (x) and vertical (y) motions must be separated and analyzed independently.

0

0

0

In addition to the values that are true for all

projectiles, the initial vertical velocity is zero.

0yiv

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 8.9ya

yx ttt

Hint: You should always list your known values (preferably in the picture when applicable) at the beginning of any problem and assign those values variables.

And for horizontal projectiles… 0yiv

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 8.9ya

yx

yi

ttt

v

0

Add the given values to our list of known values. Now that the diagram is drawn and labeled and we have identified and listed all of our “known” and “given” values for the problem, let’s begin by finding time.

Knowns:Givens:

m35

m/s 5

d

v

y

xi

A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, 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 8.9ya

yx

yiy

ttt

vv

0top

Since we know more values for vertical motion, let’s use it to find time. Start with the displacement equation (can use down as positive since all are down)…

Knowns:Givens:

m35

m/s 5

d

v

y

x

22

1

22

1

)8.9(035 t

attvd yiy

Now solve for t… sec67.29.4

35

9.435 2

t

t


Vx= 5 m/s

dy=35 m

Range, dx

0

constant

x

x

a

v

2m/s 81.9ya

yx

yi

ttt

v

0

Now that we know time, let’s find dx. Remember that horizontal motion is constant. Let’s use the displacement formula again but in the x direction.

Knowns:Givens:

m35

m/s 5

d

v

y

x

st 67.2Calculated:

m 4.13s 67.2m/s 5

22

1

x

xxix

d

tatvd 0


Vx= 5 m/s

dy=35 m

Range, dx

0

constant

x

x

a

v

2m/s 81.9ya

yx

yi

ttt

v

0

Final velocity requires a little more thought. Remember that 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.

Knowns:Givens:

m35

m/s 5

d

v

y

x

st 67.2Calculated:

m 4.13xd

Vf

θ


Vx= 5 m/s

dy=35 m

Range, dx

0

constant

x

x

a

v

2m/s 81.9ya

yx

yiy

ttt

vv

0top

Let’s look more closely at the vector, vf. To help see it better, let’s exaggerate the angle. Since x- and y- motion are separate, there must be components.

Knowns:Givens:

m35

m/s 5

d

v

y

x

st 67.2Calculated:

m 36.13xdVf

Vf

Vxf = 5 m/s

Vyf

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

θ

5 /xi xfv v m s


Vx= 5 m/s

dy=35 m

Range, dx

0

constant

x

x

a

v

2m/s 8.9ya

yx

yi

ttt

v

0

To 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.

Knowns:Givens:

m35

m/s 5

d

v

y

x

st 67.2Calculated:

m 4.13xd

Vf

Vf

Vxf = 5 m/s

Vfy

θ

m/s 2.26

)67.2)(m/s 8.9(0 2

yf

yf

yiyf

v

sv

atvv

Calculating vyf:

Still not finished. Must put components together for final velocity.

θ


Vx= 5 m/s

dy=35 m

Range, dx

0

constant

x

x

a

v

2m/s 8.9ya

yx

yiy

ttt

vv

0top

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.

Knowns:Givens:

m35

m/s 5

d

v

y

x

st 67.2Calculated:

m 4.13xd

Vf

Vf

θ

θ

2.79tan

(speed) magnitude m/s 7.26

2.265

52.261

2222

f

yfxff

v

vvv

Putting it together to calculate vf:

Final Velocity = 26.7 m/s, 79.2° below horizontal

Vxf = 5 m/s

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.

vyi

θ

vxi

vix

vi

sin

cos

iyi

ixi

vv

vv

Now that we have the components

of the initial velocity, we will use only those for calculations. **Never use the original velocity at the angle in an equation!

viy

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. So the angle of launch and angle of landing will be equal and the magnitude of the initial and final velocity will be equal. Therefore the magnitudes of the components will also be equal.

xf

yfv

v

yfxff vvv

1

22

tan

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 (in free-fall), the initial and final y-components are equal but opposite.

yiyf vv

vxf

vyfvf

vxf

vxf vyf

vyf

vf

vf

θθ

θ

vi

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

Knowns (for all projectiles):

0

(constant) vxf

x

xix

a

vv

2m/s 8.9ya

yiyf

yx

y

vv

ttt

v

0top

Vy top = 0

Vyi

Vxi

dymax =height

Range, dx

v xf = v xi

Vyf = -v yivf

vi

θ


0

constant

x

x

a

v

2m/s 8.9ya

yx

y

ttt

v

0top

Vx

Vy top = 0

Vy dymax =height

Range, dx

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

sin

cos

iyi

ixi

vv

vv

Step 3: Find 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.

vi

θ

Vy top = 0

Vy

Vx

dymax =height

Range, dx

Find time – Method 2: Use vertical motion and symmetry. Remember that the y component of initial and final velocities are equal and opposite. So using the formula

, plug in the known values from symmetry , vfy=-viy and solve for t.

vf

θvfx

vfy

tavv yyiyi

y

yi

yyiyi

av

t

tavv

2

Note: usually any vectors acting upward such as initial velocity are considered to be positive, therefore acceleration due to gravity is negative (-9.8) so the time will NOT turn out to be a negative.

vi

θ

Vy top = 0

Vy

Vx

dymax =height

Range, dxFind time – Method 3: Remember for all projectiles, the vertical velocity at the very top of the path is zero. If the projectile is launched and lands at the same height, the top of the path occurs at exactly the half-way point. If we use this as either the initial or final velocity we can calculate ½ the time of flight and then simply double it! Beginning with the same equation,

vf

θvfx

vfy

atvv yiyf

8.92

0

yiup

yi

vtort

atv Remember to use the symmetry between initial and final y-velocities!

To work from launch to top of path…

OR…To work from top of path to landing …

8.9

0

0

yidown

yyi

yyf

vt

tav

tav

Remember: If you want hang time, double tup!

V i= 3

0 m/s

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


0

constant

x

x

a

v

2m/s 8.9ya

yx

y

ttt

v

0top

Vy top = 0

Vyi

Vxi

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.


0

constant

x

x

a

v

2m/s 8.9ya

yx

y

ttt

v

0top

Vy top = 0

Viy

Vx

dymax =height

Range, dx


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 – Method 1: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

9.4260

)8.9(260

9.426

2

22

1

22

1

t

tt

tt

tatvd yyiy


vix =15 m/s

Since t = 0, could cancel a “t” or ignore that solution.

vi

θ

Vy top = 0

dymax =height

Range, dx

Find time – Method 2: Remember that the y component of initial and final velocities are equal and opposite. So, to calculate…

vf

θ

sec 3.58.952

)8.9(2626

t

t

tavv yyiyf


v y =

26 m

/sv

fy =-26 m

/s

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

Note: This method gives the same answer as the previous one. Choose the one that make the most sense to you and use it.

vi

θ

Vy top = 0

dymax =height

Range, dxFind time – Method 3: Remember for all projectiles, the vertical velocity at the very top of the path is zero. If the projectile is launched and lands at the same height, the top of the path occurs at exactly the half-way point.

vf

θ

sec 3.5)2(65.2

sec 65.2

)8.9(260

8.926

t

t

t

tavv

up

up

yyiyfTo work from launch to top of path…

OR…To work from top of path to landing…

v 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

sec 3.5)2(65.2

sec 65.2

)8.9(026

81.926

t

t

t

tavv

down

down

yyiyf

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…


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


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


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 because it is launched and lands at the same level, maximum height occurs exactly half-way through the flight. So…USE ½ of the total time of flight.

vf

θv y =

26 m

/sv

y =-26 m


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

m 5.34

))( 8.9())( 26( 22

s 3.5s

m2

12

s 3.5

22

1

2

y

sm

y

yyiy

d

d

tatvdMethod 1:

Note: If you use the top as initial point and the end as final point, vyi = 0. Then dy = - 34.5 m but height = 34.5 m.

vi

θ

Vy top = 0

dymax =height

Range, dx

To find maximum height for this problem, you must use ½ of the total time of flight and final vertical velocity of 0 if initial point is at the beginning and final point is at the top.

vf

θv y =

26 m

/sv

y =-26 m


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

Method 2:

Note: If you use the top as initial point and the end as final point, vyi = 0 and v yf = -26 m/s . Then dy = - 34.5 m but height = 34.5 m.

2

26 0 5.3

2 2

34.5

yi yfy

y

y

v vd t

d

d m

vi

θ

Vy top = 0

dymax =height

Range, dx

To find maximum height for this problem, you must use final vertical velocity of 0 if initial point is at the beginning and final point is at the top. Time is not needed for this equation!

vf

θv y =

26 m

/sv

y =-26 m


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

Method 3:

Note: If you use the top as initial point and the end as final point, vyi = 0 and v yf = -26 m/s . Then dy = - 34.5 m but height = 34.5 m.

2 2

2 2

2

(0) (26) 2( 9.8)( )

34.5

yf yi y y

y

y

v v a d

d

d m

022

1 tatvd yyiy

Vertical displacement is not zero. Consider the launch point as the zero height and then vertical displacement, dy, will be a positive number (as long as you continue to assume up is positive. Plug the value for dy into the above equation and solve for time. (Hint: Graph it and find the zeros! It’s easier than the quadratic equation.) The answer must be positive. This will be the time for the flight to that point. Then you can use that time to find the horizontal distance the object traveled to get to that point.

22

1 tatvd xxix

viθ

dy

dx

0NOTE: The highest point of the projectile DOES NOT occur at the half-way point of the flight. BE CAREFUL!

vi

θ

-dy

dx

Again, vertical displacement is not zero. If the launch point as the zero height, vertical displacement, dy, will be a negative number (as long as you continue to assume up is positive. Plug the value for dy into the distance equation and solve for time. This will be the time for the flight to that point. Then you can use that time to find the horizontal distance the object traveled to get to that point. NOTE: The highest point of the projectile DOES NOT occur at the half-way point of the flight. BE CAREFUL!2

21 tatvd xxix

0

022

1 tatvd yyiy

vi

θi

dx

The key to this problem is drawing the picture and finding the components. Again, vertical displacement is negative along with the vertical component of the initial velocity and acceleration. Remember, to find final velocity, you must find the final y- velocity and then put the x- and y- components of the final velocity back together using the pythagorean theorem and the inverse tangent.

Hints for projectiles that are thrown downward

-dy

022

1 tatvd yyiy

22

1 tatvd xxix 0

vix

viy

vfy

vfx

vf

θf

Some important notes: The horizontal velocity remains the same (drawing shows otherwise but it is not to scale). The angle is is thrown downward is not the same as the angle it strikes the ground. Finally, the projectile will never strike perpendicular to the ground!

Documents

Projectile Motion