Projectiles at an Angle

Hardest thing yet.

A progression:A) Freefall Review:

A ball drops for 3 s. How far does it fall? How fast is it going after 3 s?

vi = 0 m/st = 3sa = -10 m/s2

d = ?d = vit + ½ at2 d = ½ at2

d = ½ (-10m/s2)(3s)2 d = ½ (-10m/s2)(9s2) d = 45 m

vf = vi + atvf = atvf = (-10m/s2)(3s)

vf = 30 m/s0

0

Next…B) Throwing upward:

A ball is thrown upward with an initial velocity of 30 m/s. How far does the ball travel up in the air? How long before it hits the ground?

•Divide the motion into two halves- up and down

•Symmetry:timeup = timedown

vi up = -vf down

dup = - ddown

vi up = -vf down

30 m/s = -30 m/sa = -10 m/s2

vi = 0 m/s

vf2 = vi2 + 2ad0

2a 2ad = vf2

2ad = (-30 m/s)2

2(-10 m/s2)d = (900 m2/s2) (-20 m/s2)d = 45 m

vi up = -vf down

30 m/s = -30 m/sa = -10 m/s2

vi = 0 m/sd = 45 m

t=?vf = vi + atvf = at

0

aat = vf at = -30 m/s -10 m/s2

t = 3 s

Next…C) Horizontal Projectiles:A balls rolls off a table with a velocity of 6 m/s and lands 2 m away from the table. How long did it take the ball to hit the ground? How high is the table?

x yvix = 6m/s viy = 0m/s

vfx = 6m/s vfy =

ax = 0m/s2 ay = -10m/s2

dx = 2m dy = .45m

time = .3 s

The hardest…D) Projectiles launched at an angle:

A ball is launched with a velocity of 100 m/s at an angle of 17.4576º above the horizon. How long did it take for the ball to hit the ground? What is the range of the ball? What was the maximum height of the ball?

θ

Range

Height

• Our projectile is no longer launched horizontally so it no longer has an viy= 0 m/s

• The vi is divided between the x and y directions- we need to do something called “Resolving the x and y components”

θ = 17.4576º

Range

Height

Vi = 100m/s

• Basically the vi given at a certain angle is the resultant velocity of the horizontal (x) and vertical (y) velocity vectors.

• We have to work backwards.• We need to use trigonometry.

y

x

Before: Now:

Trigonometry- finding the side or angles of a triangle

SOHCAHTOA

v i = 100 m/s

vix

viy h

o

aθθ = 17.4576º

sin Ɵ = o h

cos Ɵ = a h

tan Ɵ = o a

viy

Let’s find vix (a) knowing vi (h):

Vi = 100 m/s

vix

viy h

o

aθθ = 17.4576º

cos Ɵ = a h

cos Ɵ = vix

vi

vi vi

vix = vi cos Ɵ

Let’s find viy (o) knowing vi (h):

v i = 100 m/s

vix

viy h

o

aθθ = 17.4576º

sin Ɵ = o h

sin Ɵ = viy

vi

vi vi

viy = vi sin Ɵ

Let’s use these new formulas to find the vix and viy for our example:

vix = vi cos Ɵ

vix = 100 m/s cos 17.4576°

vix = 100 m/s (.95)

vix = 95 m/s

viy = vi sin Ɵ

v i = 100 m/s

vix

viy θ = 17.4576º

viy = 100 m/s sin 17.4576°

viy = 100 m/s (.3)

viy = 30 m/s

vix = ? viy = ?

Stop and Check:

100 m/s30 m/s

95 m/sθ

a2 + b2 = c2

(30 m/s)2 + (95 m/s)2 = c2

(900 m2/s2) + (9025 m2/s2) = c2

√9925 m2/s2 = c~100 m/s = c

Now we can take this information and put it into a new form of our table:

vi =

θ = x y

vix =

vfx =

ax =

dx =

viy =

vfy =

ay =

dy =

½ t = t =

•Don’t forget about symmetry:timeup = timedown

vi up = -vf down

dup = - ddown

Revisit our problem:

A ball is launched with a velocity of 100 m/s at 17.4576º above the horizon. How long did it take for the ball to hit the ground? What is the range of the ball? What was the maximum height of the ball?

vi = 100 m/s

θ = 17.4576ºx y

vix = 95 m/s

vfx = 95 m/s

ax = 0 m/s

dx =

viy = 30 m/s

vfy = -30 m/s

ay = -10 m/s2

dy =

½ t = t =

viy = 30 m/s

vix = 95 m/s

Vi = 100 m/sviy = 30 m/s

1. Find the maximum height (dy):

θ = 17.4576º

Range

HeightVi =

100m

/s

HEIGHT WILL LEAD YOU TO TIME

Height is a vertical variable- Need to use info from the y direction only:

vi = 100 m/s

θ = 17.4576ºx y

vix = 95 m/s

vfx = 95 m/s

ax = 0 m/s

dx =

viy = 30 m/s

vfy = -30 m/s

ay = -10 m/s2

dy =

½ t = t =

Can we use:vfy

2 = viy2 + 2aydy

We can only use this equation if we analyze just half of the projectiles motion and make it like a

drop problem:viy = 0 m/s

vfy2 = viy

2 + 2aydy

dy = ?

θ = 17.4576º

Range

Vi = 100m/sVtop =

Vf =

θ = 17.4576º

Range

Vi = 100m/sVtop =

Vf =

For time also just analyze half the motion. Now this becomes like a horizontal projectile problem.

Remember tup = tdown

2. Finding time:

viy = 0 m/s

dy = viyt + ½ ayt2

t = ?vi = 100 m/s

θ = 17.4576ºx y

vix = 95 m/s

vfx = 95 m/s

ax = 0 m/s

dx =

viy = 30 m/s

vfy = -30 m/s

ay = -10 m/s2

dy = 45 m

½ t = t =

3. Finding the range (dx):

dx = vixt + ½ axt2

dx = ?vi = 100 m/s

θ = 17.4576ºx y

vix = 95 m/s

vfx = 95 m/s

ax = 0 m/s

dx =

viy = 30 m/s

vfy = -30 m/s

ay = -10 m/s2

dy = 45 m

½ t = 3s

t = 6s

vi = 100 m/s

θ = 17.4576ºx y

vix = 95 m/s

vfx = 95 m/s

ax = 0 m/s

dx = 570 m

viy = 30 m/s

vfy = -30 m/s

ay = -10 m/s2

dy = 45 m

½ t = 3s

t = 6s

Completed Chart

v i = 100 m/s

Θ = 17.4576°

4. Finding components using a ruler and a protractor:• Used when the #s aren’t easy and you

don’t have a calculator• 1st thing to do is make an x-y axis

using your protractor and ruler that is a 90°

• Set a scale- 1 cm = 10 m/s• Draw in the vi vector at the correct

scaled length with the correct angle• Draw in the viy line- be careful!!! The

length of this line, converted using the established scale, is the magnitude of the initial velocity in the y direction (for our example it should be 3 cm)

• Draw in the vix line- The length of this line, converted using the established scale, is the magnitude of the initial velocity in the x direction (for our example it should be 9.5 cm)

v i = 100 m/s = 10 cm

17.4576°vix = 95 m/s = 9.5 cm

viy = 30 m/s = 3 cm

sin cos tan0° 0 1 0

30° .5 .866 .577

45° .707 .707 1

60° .866 .5 1.73

90° 1 0 ____

Common sin/cos/tan to MEMORIZE

HW #1:A ball is launched with a velocity of 100 m/s at 30º above the horizon. How long did it take for the ball to hit the ground? What is the range of the ball? What was the maximum height of the ball?

θ

Range

Height

Let’s use these new formulas to find the vix and viy for our example:

vix = vi cos Ɵ

vix = 100 m/s cos 30°

vix = 100 m/s (.866)

vix = 86.6 m/s

viy = vi sin Ɵ

v i = 100 m/s

vix

viy θ = 30º

viy = 100 m/s sin 30°

viy = 100 m/s (.5)

viy = 50 m/s

vix = ? viy = ?

Stop and Check:

100 m/s50 m/s

86.6 m/sθ

a2 + b2 = c2

(50 m/s)2 + (86.6 m/s)2 = c2

(2500 m2/s2) + (7500 m2/s2) = c2

√10000 m2/s2 = c100 m/s = c

Revisit our problem:

A ball is launched with a velocity of 100 m/s at 30º above the horizon. How long did it take for the ball to hit the ground? What is the range of the ball? What was the maximum height of the ball?

vi = 100 m/s

θ = 30ºx y

vix =86.6 m/s

vfx =86.6 m/s

ax = 0 m/s

dx =

viy = 50 m/s

vfy = -50 m/s

ay = -10 m/s2

dy =

½ t = t =

viy = 50 m/s

vix = 86.6 m/s

Vi = 100 m/sviy = 50 m/s

1. Find the maximum height (dy):

θ = 30º

Range

HeightVi =

100m

/s

HEIGHT WILL LEAD YOU TO TIME

vi = 100 m/s

θ = 30ºx y

vix =86.6 m/s

vfx =86.6 m/s

ax = 0 m/s

dx =

viy = 50 m/s

vfy = -50 m/s

ay = -10 m/s2

dy =

½ t = t =

Make it like a drop problem:viy = 0 m/s

vfy2 = viy

2 + 2aydy

dy = ?

Height is a vertical variable- Need to use info from the y direction only:

θ = 30º

Range

Vi = 100m/sVtop =

Vf =

Now this becomes like a horizontal projectile problem.

Find time.

Remember tup = tdown

2. Finding time:

vi = 100 m/s

θ = 30ºx y

vix =86.6 m/s

vfx =86.6 m/s

ax = 0 m/s

dx =

viy = 50 m/s

vfy = -50 m/s

ay = -10 m/s2

dy = 125 m

½ t = t =

viy = 0 m/s

dy = viyt + ½ ayt2

t = ?

3. Finding the range (dx):

dx = vixt + ½ axt2

vi = 100 m/s

θ = 30ºx y

vix =86.6 m/s

vfx =86.6 m/s

ax = 0 m/s

dx =

viy = 50 m/s

vfy = -50 m/s

ay = -10 m/s2

dy = 125 m

½ t = 5s

t = 10s

dx = ?

4. Finding components using a ruler and a protractor:• Used when the #s aren’t easy and you

don’t have a calculator• 1st thing to do is make an x-y axis

using your protractor and ruler that is a 90°

• Set a scale- 1 cm = 10 m/s• Draw in the vi vector at the correct

scaled length with the correct angle• Draw in the viy line- be careful!!! The

length of this line, converted using the established scale, is the magnitude of the initial velocity in the y direction (for our example it should be 5 cm)

• Draw in the vix line- The length of this line, converted using the established scale, is the magnitude of the initial velocity in the x direction (for our example it should be 8.6 cm)

v i = 100 m/s = 10 cm

30°vix = 86.6 m/s = 8.6 cm

viy = 50 m/s = 5 cm

vi = 200 m/s

θ = 20.4873º

x y

vix =

vfx =

ax =

dx =

viy =

vfy =

ay =

dy =

½ t = t =

HW #2-

v i = 200 m/s

Θ = 20.4873°

cos of 20.4873 = .94sin of 20.4873 = .35

Set up your paper like this-

v i = 200 m/s

Θ = 20.4873°

vi = 200 m/s

θ = 20.4873º

x y

vix =

vfx =

ax =

dx =

viy =

vfy =

ay =

dy =

½ t = t =

Resolving Components:

dy=?

dx=?

t=?

FRONT BACK

v i = 200 m/s = 10 cm

20.4873°vix = ? cm = ? m/s

viy = ? cm = ? m/s

Answer:vi = 200 m/s

θ = 20.4873º

x yvix = 188m/s

vfx = 188m/s

ax = 0m/s2

dx = 2632m

viy = 70m/s

vfy = -70m/s

ay = -10m/s2

dy = 245m

½ t = 7st = 14s

cos Ɵ = a h

cos Ɵ = vix

vi

vi vi

vix = vi cos Ɵ

viy = vi sin Ɵ

vix = 200 m/s cos 20.4873°

vix = 200 m/s (.94)

vix = 188 m/s

viy = 200 m/s sin 20.4873°

viy = 200 m/s (.35)

viy = 70 m/s

HW #3

dy = 125 m

dX = 180 m

Orange Review Book

v i = 200 m/s

vix

viy θ = 30º

vix = vi cos Ɵ

viy = vi sin Ɵ

v i = 25 m/s

vix

viy θ = 53º

vix = vi cos Ɵ

viy = vi sin Ɵ

v i = 15 m/s

vix

viy θ = 35º

vix = vi cos Ɵ

viy = vi sin Ɵ

v i = 60 m/s

vix

viy θ = 37º

vix = vi cos Ɵ

viy = vi sin Ɵ

v i = 19.6 m/s

vix

viy θ = 30º

vix = vi cos Ɵ

viy = vi sin Ɵ

vi = 19.6 m/s

θ = 30ºx y

vix = m/s

vfx = m/s

ax = 0 m/s

dx =

viy = m/s

vfy = - m/s

ay = -10 m/s2

dy =

½ t = t =

viy = vi sin Ɵ

vi = 40 m/s

θ = 40ºx y

vix = m/s

vfx = m/s

ax = 0 m/s

dx =

viy = m/s

vfy = - m/s

ay = -10 m/s2

dy =

½ t = t =

vix = vi cos Ɵ

vix = vi cos Ɵ

vi = 8.5 m/s

θ = 35ºx y

vix = m/s

vfx = m/s

ax = 0 m/s

dx =

viy = 4.9m/s

vfy = - 4.9m/s

ay = -10 m/s2

dy =

½ t = t =

vi = 40 m/s

θ = 60ºx y

vix = m/s

vfx = m/s

ax = 0 m/s

dx =

viy = m/s

vfy = - m/s

ay = -10 m/s2

dy =

½ t = t = 7.1 s

vi = m/s

θ = ºx y

vix = 49 m/s

vfx = 49 m/s

ax = 0 m/s

dx =

viy = 98 m/s

vfy = -98 m/s

ay = -10 m/s2

dy =

½ t = t =

v i = m/s

vix

viy θ = º

vix = 49 m/s

viy = 98 m/s

vi = m/s

θ = ºx y

vix = 15 m/s

vfx = 15 m/s

ax = 0 m/s

dx =

viy = 25 m/s

vfy = -25 m/s

ay = -10 m/s2

dy =

½ t = t = 5 s

Which launch angle gives the greatest range?

• The cannonball launched at a 45-degree angle had the greatest range. The cannonball launched at a 60-degree angle had the highest peak height before falling. The cannonball launched at the 30-degree angle reached the ground first.

vi =

θ = x y

vix =

vfx =

ax = 0 m/s2

dx =

viy =vfy =ay = -10 m/s2

dy =

½ t = t = 8s

HW #4

vi = ? m/sΘ = ?°

vix = 5 m/s

Must use protractor/ruler

• 1st thing to do is make an x-y axis using your protractor and ruler that is a 90°• Set a scale- 1 cm = 2.5 m/s• Draw in the vectors at the correct scaled length with the correct angle

v i =°

vix = 5 m/s = 2cm

v iy =

40 m

/s=

16cm

v i =°

vix = 5 m/s = 2cm

v iy =

40 m

/s=

16cm

vi = 16.1 cm = 40.25 m/s83°

vix = 5 m/s = 2cm

v iy =

40 m

/s=

16cm

vi =

θ = 70°x y

vix =

vfx =

ax = 0 m/s2

dx =

viy = 100 m/s

vfy =ay = -10 m/s2

dy =

½ t = t =

HW#5

viy = 100 m/s

Θ = 70°

• 1st thing to do is make an x-y axis using your protractor and ruler that is a 90°• Set a scale- 1 cm = 10 m/s• Draw in the vectors at the correct scaled length with the correct angle

70°

v iy =

100

m/s

= 10

cm

vi

v iy =

100

m/s

= 10

cm

vi

v iy =

100

m/s

= 10

cm

vix

vi = 106 m/s

θ = 70°x y

vix = 35 m/s

vfx = 35 m/s

ax = 0 m/s2

dx =

viy = 100 m/s

vfy = -100m/s

ay = -10 m/s2

dy =

½ t = t =

HW#5

viy = 100 m/s

Θ = 70°

Find in order:1.dy

2.1/2t 3.t4.dx

dy =? “Drop”= viy = 0m/s

vfy2 = viy

2 + 2aydy

2ay 2ay

dy = vf2

2ady = (100 m/s)2

2(-10 m/s2)dy = (10000 m2/s2) (-20 m/s2)dy = 500 m

0

½ t=? “Drop”= viy = 0m/s

t=?vf = vi + atvf = at

0

aat = vf at = -100 m/s -10 m/s2

1/2t = 10 s

t = 20 s

dx=?

vix = dx

ttt

dx = vix tdx = 35 m/s (20s)dx = 700 m

dx = vixt + 1/2axt0

dx = vix tdx = 35 m/s (20s)dx = 700 m

vi = 106 m/s

θ = 70°x y

vix = 35 m/s

vfx = 35 m/s

ax = 0 m/s2

dx = 700m

viy = 100 m/s

vfy = -100m/s

ay = -10 m/s2

dy = 500 m

½ t = 10s t = 20s

HW#5- Complete

viy = 100 m/s

Θ = 70°

Practice Problem:

A golf ball is hit with a velocity of 20 m/s at 45 above the horizon. Find the time it takes the golf ball to travel, the range of the golf ball, and the maximum height the golf ball reaches.

A player kicks a football from ground level at 27.0 m/s at an angle of 30.0 degrees above the horizontal.

Find:a) its “hang time” (time that the ball is in

the air), b) the distance the ball travels before it

hits the ground, and c) its maximum height.

A cannon is fired. The cannonball reaches a maximum height of 125 m. The range of the cannonball is 900 m. At what angle was the cannon fired from?

θ = ?

Range = 900 m

Height = 125 m