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

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.

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

Please ignore your eCampus grade for now!

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

Three velocity vectors.v vx vy

Three velocity vectors.

Starts with horizontal velocity vx0

…And vertical velocity vy0 = 0 m/s.

v vx vy

v = vx0

Free body!

v = vx0

t = 0s

ay = -9.8 m/s2

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

• 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

Free body!

vx = vx0

t = 0s

t = 1s

t = 2s

t = 3s

v vx vy

vx = vx0

vx = vx0

vx = vx0

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

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

Generic Projectile Motion

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

• 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

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.

Q24

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.

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

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

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

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.

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)

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=0 Δx = 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).

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

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!)

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.