Projectile Motion 1

Preflight Responses• A flatbed railroad car is moving along a

track at constant velocity. A passenger riding at the center of the train throws a ball straight up. Neglecting air resistance, where will the ball land?

•     Forward of the center of the train.

•     Right at the center of the train.

•     Backward of the center of the train.

74% answered correctly!

““x” and “y” components of motion are x” and “y” components of motion are independent.independent. A man on a train tosses a ball straight up in the air.

View this from two reference frames:

Reference frame on the moving

train.

Reference frame

on the ground.

Frames of Reference

Projectile: An Object Moving Solely Under Influence of

Gravity

Preflight ResponsesYou and a friend are standing on level ground, each holding identical baseballs. At exactly the same time, and from the same height, you drop your baseball without throwing it while your friend throws her baseball horizontally as hard as she can. Which ball hits the ground first?      Your ball

     Your friends ball      They both hit the ground at the same time.

79% answered correctly!

Horizontally Launched Projectile

Projectile Motion

Without gravity, the cannonball would follow a straight-line path as shown in the picture.

With gravity, the cannonball falls the same distance below the gravity-free path as a cannonball dropped from the same height.

Motion in Freefall

HorizontalMotion

VerticalMotion

Acceleration

NO

Velocity Constant

Motion in Freefall

HorizontalMotion

VerticalMotion

Acceleration

NO

YES – “g” is downward at

9.8 m/s/s

Velocity Constant

Changing by

9.8 m/s, downward

each second

Independence of x & y Motions

t (s) Vy(m/s) Y (m)

0

1

2

3

4

t

X (m)

0

5

10

15

20

vxt

For Vx= 5 m/s

Independence of x & y Motions

t (s) Vy(m/s) Y (m)

0 0

1 9.8

2 19.6

3 29.4

4 39.2

t 9.8 t

X (m)

0

5

10

15

20

vxt

For Vx= 5 m/s

Independence of x & y Motions

t (s) Vy(m/s) Y (m)

0 0 0

1 9.8 4.9

2 19.6 19.6

3 29.4 44.1

4 39.2 78.4

t 10 t 4.9t2

X (m)

0

5

10

15

20

vxt

For Vx= 5 m/s

Problem:

• The pilot of a hovering helicopter drops a lead brick from a height of 1000 m. How long does it take to reach the ground and how fast is it moving when it gets there? (neglect air

resistance)

1000 m

Problem:• First choose coordinate

system. --Origin and y-direction

• Next write down position equation:

Realize that

20 0y

1y y v t at

2

21

2y gt

1000 m

y = 1000 m

y

0, 0,yo ov y a g

Problem:• Solve for time t when y = 1000 m

•

• To find the velocity:

• Evaluate vy:

y0 = 1000 m

2

2 2 100014.3

9.8

y mt s

g m s

21

2y gt

y = 1000 m

yv gt

(9.8 / / )(14.3 )

140 /

yv m s s s

m s

y

Problem 2:

• A helicopter is traveling toward the west at 30 m/s. The pilot of drops a lead brick from a height of 1000 m. How long does it take to reach the ground and what is its velocity when it gets there? How far west of the release point does the brick hit the ground?

(neglect air resistance)

1000 m

Problem 2:

1000 m

2

2 2 100014.3

9.8

y mt s

g m s

y

x

Find the time as before – using the :vertical motion equation:

Since the time to reach the ground is INDEPENDENT OF HORIZONTAL MOTION

x

Find the horizontal position (x) using the constant velocity equation for the horizontalmotion equation:

(30 )(14.3 )

429

mx sx v t s

x m

Problem 2:

1000 m

y

x

The y-velocity, from before is:

To find the speed when it reaches the ground, we must use vector addition to combine the x and y velocities.

x

The x-velocity is constant

2 22 2 30 140

143

m ms sx y

ms

v v v

v

(9.8 / / )(14.3 )

140 /

yv m s s s

m s

30 mx sv

1 140tan 77.930

mos

ms

The magnitude of the velocity is:

The angle the velocity makes with the horizontal is:

vy = 140 m/s

30 mx sv

θ

v