Vectors are quantities with Magnitude AND Direction

Ex:› Displacement› Velocity› Acceleration› Force

Scalars are quantities with only magnitude

Ex:› Distance› Speed› Time› Mass

Vector notation:

The length of the arrow represents magnitude

The direction of the vector, found from the angle, represents direction

θ

A2 + B2 = C2

A surveyor stands on a riverbank directly across the river from a tree on the opposite bank. She then walks 100 m downstream, and determines that the angle from her new position to the tree on the opposite bank is 50°. How wide is the river, and how far is she from the tree in her new location?

You are standing at the very top of a tower and notice that in order to see a manhole cover on the ground 50 meters from the base of the tower, you must look down at an angle 75° below the horizontal. If you are 1.80 m tall, how high is the tower?

Projectiles

a. A projectile is any object that falls through the air.

b. Its path is called a trajectory.c. Forces acting on projectiles are:

i. Gravityii. Air Resistance (which we’ll ignore

because it’s small)

a. The projectile’s velocity must be broken into x- and y- components

b. Vectors that are perpendicular to each other act independently!

b. Vectors that are perpendicular to each other act independently

c. Horizontal components and vertical components don’t affect each other!

d. We still have to make a few assumptions:

i. g = 9.81 m/s2 (down)ii. air resistance is

negligableiii. no horizontal aiv. rotation of Earth can

be ignoredv. Time is the same for

each component

e. The velocity of a projectile has two components, vx and vy.

i. vy = v sin

ii. vx = v cos

v

vx

vy

Find the x- and y-components of the following vectors

R = 175 meters @ 95o v = 25 m/s @ -78o a = 2.23 m/s2 @ 150o

Remember: horizontal velocity is always CONSTANT! There is NO acceleration in the x-direction! Acceleration in the y-direction is due to gravity

co n sta n t h o rizo n ta l ve loc ityvx

ve rtica l ve loc ity in c rea se s w ith tim e vy

vx

a. A projectile launched upward at an angle would have a parabolic path

b. It moves horizontally at constant speedc. It accelerates downward at a rate of g

= 9.81 m/s2

1. A flagpole ornament falls off the top of a 25.0 m flagpole. How long would it take to hit the ground?

2. A stone is thrown horizontally from the top of a cliff that is 44.0 m high. It has a horizontal velocity of 15.0 m/s. We want to find how long it takes the stone to fall to the deck and how far it will travel from the base of the cliff.

Tip: break each problem into two separate problems

We assume that the ornament has no horizontal velocity.

Plug in givens:

21

2y gt 2 2y

tg

2yt

g

2

2 25.02.26

9.80

mt s

m

s

This is like the flagpole problem, except that the stone has an initial horizontal velocity. But we know that the time it takes to hit the ground is the same as if it were falling straight down. (This is the key concept!!!)

21

2y gt

2

2 40.022.86

9.80

myt s

mgs

Now that we know how long it takes to fall, we can figure out the horizontal distance it travels before it hits the ground. It has a constant horizontal speed, and can travel sideways at this speed as long as it is in the air falling, so to find x we use its average velocity and the time:

15.0 2.86 42.9x xx m

v x v t s mt s

A B-17 (a World War II era multiengine bomber) is flying at 375 km/h. The bombs it drops travel a horizontal distance of 5 250 m. What was the altitude of the plane at the time they had the old "bombs away"?

We have to find the time for the bomb to travel a horizontal distance of 5 250 m:

First we convert the bomber’s speed to meters per second:

Then we can find the time it takes the bomb to travel a horizontal distance of 5 250 m:

1000 1375 104.2

1 3600

km m h m

h km s s

15 250 50.4

104.2x

x

x xv t m s

mt vs

Now we can find the vertical distance (altitude):

222

1 19.80 50.4 12 400

2 2

my at s m

s

a. A projectile (with no air resistance) will always have a horizontal velocity that remains constant.

b. If it is shot upward, gravity will slow down its vertical velocity

c. At the top of the path, vy = 0

d. At the bottom of the arc, final vertical velocity = initial vertical velocity in opposite directions.

e. The projectile travels a horizontal distance of x, also called the range.

A ball is given an initial velocity of 22. 7 m/s at an angle of 66.0 to the horizontal. Find how high the ball will go.

To solve this problem, we have to find the vertical velocity of the ball. Once we know it, we can find how high it goes.

osin 22.7 sin 66.0 20.7ym m

v vs s

The ball starts out with vy and rises till its vertical velocity is zero. We can use these as the initial and final velocity of the ball.

Note that we have to pay attention here to the sign of the motion. We have both down and up motion and have to be clear about which direction is possible. Of course if the motion is in only one direction, we don’t have to worry about it.

22 2 2 22 0 2 2

2o

y o o ov

v v ay v ay v ay ya

2

2

20.721.9

2 9.80

ms

y mm

s

First find the vertical velocity:

Now we can find the time:

osin 345 sin 32.0 182.8ym m

v vs s

y ov v at o yv v

Now we can find the range since we know the time. First we find the horizontal velocity:

y oy o

v vat v v t

g

2

182.8 182.837.31

9.80

m ms st s

m

s

o0cos 345 cos32.0 292.6x

m mv v

s s

292.6 37.31 10 900x xx m

v x v t s mt s

a. If you’re better at memorization than you are at math, well you probably should’ve chosen a different science, but here are a couple helpful formulas:

1. Range:

2. ymax: