Projectile MotionGood practice problems in book:

3.23, 3.25, 3.27, 3.29, 3.31, 3.33, 3.43, 3.47, 3.51, 3.53, 3.55

a = -9.8 m/s2

v

A projectile is anything experiencing free-fall, particularly in two dimensions.3.23, 3.25, 3.27, 3.29, 3.31, 3.33, 3.43, 3.47, 3.51, 3.53, 3.55

Practice testReminder: test Feb 8, 7-10pm!Email me if you have conflicts!

Treat this practice test like the exam.3 hours, closed book, non-graphing

calculator, pencil and paper.

This is your next homework (not graded).

Do it ASAP - it will indicate how well you’ll probably do on the exam.

I’m not going to ask you to turn it in. While you guys are passing these around I’m going to give a few announcements.

Your intuitive understanding of the Physical world

• Note: all of you got an A.• Tested your perception of basic Physics principles.• We’re going to work on fixing your misconceptions!• For instance: everything falls at the same rate!

Percent grasp of general Physics principles

Forget for a moment that I’m grading you

Please ignore your eCampus grade for now!

Please ignore the grades posted there until I get things fixed. They might not make sense.

Today!

• Decomposing X and Y movements.

• How to hit a target.

• How to go about Projectile Motion problems.

Figure 1. Dysfunctional

projectile

Q22

A B CStraight down after some time

Never completely straight down

Ignoring air resistance, what would be the path of motion if someone ran off of a cliff?

D

Straight out for a little while then gradually

down.

Straight down almost immediately

Answer: C.Gravity always acting once in free-fall, and has forward motion with nothing pushing it back.

Three velocity vectors.v vx vy

Let’s think about this guy running off the cliff.And let’s consider that he has THREE VELOCITY VECTORS that we need to consider. There’s the net velocity he’s going in a given direction, there’s the horizontal component vx (how fast is he moving parallel to the ground), and there’s the vertical component, vy (how fast is he going toward the ground).

Three velocity vectors.

Starts with horizontal velocity vx0

…And vertical velocity vy0 = 0 m/s.

v vx vy

v = vx0

When he starts, it’s all in the horizontal component. There’s no vertical component.

Free body!

v = vx0

t = 0s

ay = -9.8 m/s2

This is just as true the very moment he leaves the cliff face. Now the moment he leaves the cliff, he becomes a free body; the only thing acting on him is the force of gravity. This means he gets an acceleration vector downward at good old 9.8m/s^2. I want you to take 30 seconds and tell me something very fundamental here.

Free body!

v = vx0

t = 0s

ay = -9.8 m/s2

Q23

Throughout this fall, what is the jumper’s horizontal acceleration

vector, ax?A. vx0/tB. 9.8 m/s2

C. 0 m/s2

D. Not given, so unknown

Answer here is C.

• Acceleration needed to change velocity

• In free bodies, the only acceleration is in the y direction

• vx is always equal to vx0 (unless it hits something)

Free body!

v = vx0

t = 0s

ay = -9.8 m/s2

vx = vx0 + axt

You need an acceleration to change your velocity. This guy, LIKE ALL PROJECTILES, is a free body. Thus, he gets affected by the force of gravity in the Y DIRECTION, this means that over the course of his journey, this jumper’s Y VELOCITY VECTOR will gradually change. This is why he falls, and falls faster and faster. However, there is nothing pushing him forward or backward in the x direction, so his horizontal velocity will never change!

Free body!

vx = vx0

t = 0s

t = 1s

t = 2s

t = 3s

v vx vy

vx = vx0

vx = vx0

vx = vx0

Let’s look at this in terms of vector decomposition of his movement. I’m showing his true velocity as the black arrowNo matter how much time goes by, velocity’s x component never changes because there’s nothing to change it! The guy keeps moving forward.But what about the y component?

t = 0svy0 = 0 m/s t = 1s

t = 2s

t = 3s

v vx vy

v2 = v02 + 2aΔx

Δx = v0t + ½ at2

v = v0 + at

How far downward has the jumper fallen

after 3 seconds?

+y

Answer: BBut what about y?Y is changing only due to gravity so we’re solving this the SAME WAY WE HAVE BEEN! Using the plain old motion equations.

t = 0s

t = 1s

t = 2s

t = 3s

v vx vy

v2 = v02 + 2aΔx

Δx = v0t + ½ at2

v = v0 + at vy = vy0 -gtΔy = vy0t -½ gt2

vy0 = 0 m/s

+y

[WROTE OUT THE Y DERIVATIONS ON THE LIGHT BOARD]By the way, this is not the ANSWER to the homework but this idea should look familiar if you’ve done the homework already!

Generic Projectile Motion

y motion: Same as vertical-only problem!x motion: Covers Δx distance at vx0 speed

This is true for ALL PROJECTILES.

• Separation of vectors into components allows separations of equations into components:

x and y: Motions are Independent

221 tatvx

tavv

xxo

xxox

+=Δ

+=2

21 tatvy

tavv

yyo

yyoy

+=Δ

+=

ax = 0 m/s2 ay = ±9.8 m/s2

The main point is here that we can treat x and y motions independently. HERE IS WHERE THE DEMO WENT TERRIBLY WRONG :).The question we pose is: I am a monkey hunter wanting to hit the monkey hanging from the tree. As soon as I shoot, the monkey lets go of the tree Think about the x and y motion of the bullet and the monkey.

Monkey Hunter

The instant the hunter fires, the monkey will reflexively let go of the tree and drop to the ground. Where should the hunter aim?

A. Above the monkey.B. At the monkey.C. Below the monkey.

Q24A classic 2D physics problem.

Monkey Hunter

If there were no gravity, the bullet would hit the monkey because the monkey would not move.

Monkey Hunter

With gravity, the bullet and the monkey fall at the same rate: the rate of the acceleration of gravity, -9.8 m/s2.

The answer is the same no matter where the hunter is standing.

(READ RED PART THEN SAY)- as long as the bullet has enough velocity to get to the monkey before it hits the ground.Another way of saying the same two things I’ve been saying:1. When solving kinematics problems, solve x and y piecewise.2. Everything falls at the same rate.

X Y

Break up what you know in terms of the horizontal and vertical…

Prevents mistakes!

“After it leaves your hand, before it hits the ground”

or t

• vy = 0 at top of trajectory • vx = vxo remains the same throughout trajectory because

there is no acceleration along the x-direction

vo = initial velocity vectorθo = initial direction of velocity vector

One other thing that’s been confusing you: the initial and final point of these problems. You’re often given a velocity and direction, and unless it notes that it’s in the X or Y direction, that v will always be v0, the initial velocity vector. That theta will be the direction of the initial velocity vector. AS TIME CHANGES, THETA WILL CHANGE AS DEFINED BY ITS X AND Y COMPONENTS. You can see that here in this figure by the angle of the primary velocity vector arrow.

Beyond this, projectile motion problems just take a lot of planning and thinking.

Take your time and think about the set-up of the problem.

What do I know?What’s the first step?What’s the next step?

Strategy for Projectile Motion Problems

Extra important to follow our problem-solving tips: Break the problem down, Draw a picture, List knowns/unknowns.

Strategy for Projectile Motion Problems

The time will be the same for x and y parts of the question.

If you don’t have enough information for x or y components, solve for time.

ΔxΔy

Here’s a far more specific and practical tip! Often times the first step is solving for time.

Throwing something off of a cliff(5 examples with increasing difficulty)

1. How much time does it take to fall?

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Throwing something off of a cliff(5 examples with increasing difficulty)

1. How much time does it take to fall?

2. How far from the base of the cliff does it hit the ground? (Need the time first)

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Throwing something off of a cliff(5 examples with increasing difficulty)

1. How much time does it take to fall?

2. How far from the base of the cliff does it hit the ground? (Need the time first)

3. How fast it is moving vertically when it hits the ground? (y component of final velocity)

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Throwing something off of a cliff(5 examples with increasing difficulty)

1. How much time does it take to fall?

2. How far from the base of the cliff does it hit the ground? (Need the time first)

3. How fast it is moving vertically when it hits the ground? (y component of final velocity)

4. What is the magnitude of its velocity when it hits the ground?

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Throwing something off of a cliff(5 examples with increasing difficulty)

1. How much time does it take to fall?

2. How far from the base of the cliff does it hit the ground? (Need the time first)

3. How fast it is moving vertically when it hits the ground? (y component of final velocity)

4. What is the magnitude of its velocity when it hits the ground?

5. What is the angle that it hits the ground from the horizontal?

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

How much time to fall? (Think Vertical)

We’re talking about something falling, and that is vertical motion, so we will only use vertical ideas and numbers.

v2 = v02 + 2aΔx

Δx = v0t + ½ at2

v = v0 + at

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Need to solve for time. Because x and y are independent, we ONLY CARE ABOUT Y HERE! That makes this problem easier.

Δy = voyt + 1/2 at2 Δy = (16 sin 60) t +1/2 (-9.8) t2

Quadratic eq. (at2+bt+c=0): a=-4.9, b=13.85, c=- Δy =15 m t=[-b ± (b2-4(a)(c))^ ½] /(2a)

=-13.85±(191.8-4(-4.9)(15))^ ½ ]/(-9.8) t= 1.41 ± 2.25= -0.84 s or 3.66 s

We’re talking about something falling, and that is vertical motion, so we will only use vertical ideas and numbers.

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

How much time to fall? (Think Vertical)

[solution from light board above]

How far from cliff base? (Think Horizontal)

v2 = v02 + 2aΔx

Δx = v0t + ½ at2

v = v0 + at

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

We know it was in the air for 3.66s (from the previous question), and it’s moving at a constant speed in the

x-direction the whole time (ax = 0).

How far from cliff base? (Think Horizontal)

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Δx = voxt+ ½ ax t2 where ax=0Δx = vx t = (16 cos60º)(3.66s)

We know it was in the air for 3.66s (from the previous question), and it’s moving at a constant speed in the

x-direction the whole time (ax = 0).

[solution from light board above]

How far from cliff base? (Think Horizontal)

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Δx = voxt+ ½ ax t2 where ax=0Δx = vx t = (16 cos60º)(3.66s)

Δx = 29m

We know it was in the air for 3.66s (from the previous question), and it’s moving at a constant speed in the

x-direction the whole time (ax = 0).

[solution from light board above]

How far from cliff base? (Think Horizontal)

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Δx = voxt+ ½ ax t2 where ax=0Δx = vx t = (16 cos60º)(3.66s)

Δx = 29m

We know it was in the air for 3.66s (from the previous question), and it’s moving at a constant speed in the

x-direction the whole time (ax = 0).

[solution from light board above]

Vertical speed when it lands? (Think Vertical)

v2 = v02 + 2aΔx

Δx = v0t + ½ at2

v = v0 + at

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Asking for y component of final velocity.It’s been accelerating down the whole time. We know that gravity is causing this acceleration, so we can figure out how fast it is going (vertically) when it hits the ground.

Vertical speed when it lands? (Think Vertical)

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Asking for y component of final velocity.It’s been accelerating down the whole time. We know that gravity is causing this acceleration, so we can figure out how fast it is going (vertically) when it hits the ground.

vfy = viy + a t vfy = 16 sin60º -9.8 x3.66

vfy = -22 m/s Definitely negative is good, since moving

opposite from positive y direction

[solution from light board above]

v2 = v02 + 2aΔx

Δx = v0t + ½ at2

v = v0 + at

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Final angle: which way is the ball going?

How do we get is its final velocity vector?

Magnitude and Angle that it hits the ground? (Finally combine!)

The equations are here as a diversion.

Magnitude and Angle that it hits the ground? (Finally combine!)

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

Vector sum!

vx

vy

Final angle: which way is the ball going?

How do we get is its final velocity vector?

v

c2 = a2 + b2

= (8.0m/s)2 + (-22m/s)2

c = 23.4 m/s

tanΘ = opp/adj = (-22 m/s) / (8m/s) Θ = tan-1(-22/8) = -70° The object is moving at 23m/s at an angle of 70° below the horizontal when it hits the ground.

-22

Magnitude and Angle that it hits the ground? (Finally combine!)

A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal.

[solution from light board above]