Adding vectors graphically

Adding vectors using the components method

Resultant vectorFind magnitude R and direction θ

Projectile Motion

Treating the vertical motion independently of the horizontal motion, and then combining them to find the trajectory, is the secret.

The vertical motion only depends on acceleration of gravity, which is the same g = 9.80 m/s2 downward acceleration, independent of the path the object takes.

t = 0.1s

t = 0.2s

t = 0.3s

t = 0.4s

t = 0.5s

t = 0.6s

Which of these three balls would hit the floor first if all three left the tabletop at the same time?

a)The ball with initial velocity v1.

b)The ball with initial velocity v2.

c)The ball with initial velocity v3.

d)They would all hit at the same time.

Since all three balls undergo the same downward acceleration, and they all start with a vertical velocity of zero, they would all fall the same distance in the same time!

This is a cool demonstration.

Problem 59 in Serway.

tvx )cos( 00 tvx )cos( 00 tvx x0 tvx x0

200 2

1)sin( gttvy 200 2

1)sin( gttvy 2

0 2

1gttvy y 2

0 2

1gttvy y

As a projectile moves in its parabolic path, the velocity and acceleration vectors are perpendicular to each other

1. everywhere along the projectile’s path.

2. at the peak of its path.

3. nowhere along its path.

4. not enough information is given.

Suppose you are carrying a ball and running at constant speed, and wish to throw the ball and catch it as it comes back down. You should

1. throw the ball at an angle of about 45° above the horizontal and maintain the same speed.

2. throw the ball straight up in the air and slow down to catch it.

3. throw the ball straight up in the air and maintain the same speed.

Quick Quizzes

What’s the general equation of a parabola?

So, why do you say the projectile motion is exactly like a parabola?

2bxaxy 2bxaxy

Because it is the combination of the horizontal motion:

with the vertical motion: 200 2

1)sin( gttvy 200 2

1)sin( gttvy

tvx )cos( 00 tvx )cos( 00

Once you combine this two equations (isolate t time from the first eq. and substitute it in the second eq.), you will get:

22

000 )cos(2)(tan x

v

gxy

2

200

0 )cos(2)(tan x

v

gxy