Projectile Motion

1 / 22

# Projectile Motion - PowerPoint PPT Presentation

Projectile Motion. Independence of vectors. When a vector is moving in two dimensions the vertical and horizontal components are independent from each other. They have no affect on each other. “Cliff Problems”. V h only. “Cliff Problems”. V h only. “Cliff Problems”. V h only. V h.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Projectile Motion' - demont

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Projectile Motion

Independence of vectors
• When a vector is moving in two dimensions the vertical and horizontal components are independent from each other.
• They have no affect on each other
“Cliff Problems
• How long object is in the air is only dependent on the height of the “cliff”
• All objects fall at the same rate, so it doesn’t matter if the horizontal velocity is 100 m/s or 0 m/s, the time in the air will be the same.
• The object’s horizontal velocity is constant, it doesn’t accelerate horizontally, only vertically
“Cliff Problems”

To find the time in the air

use the following equation:

h = gt2

2

Vh only

h

range

“Cliff Problems”

Vh only

To find the range, use the time in the air

and the following equation:

range = (Vh)(t)

h

range

“Cliff Problems”

Vh only

A cliff is 44 m high. If a car is traveling

15 m/s, how far from the base of the cliff

will it land. (Find the range)

h

range

“Cliff Problems”

Vh only

A cliff is 44 m high. If a car is traveling

15 m/s, how far from the base of the cliff

will it land. (Find the range)

• Find time in air.
• h = gt2
• 2
• 44 m = (9.8m/s2)(t2) t = 2.9 sec
• 2

h

range

“Cliff Problems”

Vh only

A cliff is 44 m high. If a car is traveling

15 m/s, how far from the base of the cliff

will it land. (Find the range)

• Calculate range:
• range = Vht
• = (15 m/s)(2.9 sec)
• = 43.5 m

h

range

“Shooting at an Angle”
• This type of problem deals with the launching of an object at an angle.
• The object has both vertical and horizontal velocities.
• These are independent of each other.
“Shooting at an Angle”
• The horizontal velocity remains constant
• The vertical velocity slows down due to gravity and then accelerates as it comes back down
• They are independent of each other
“Shooting at an Angle”

velocity

Vv

θ

Vh

A gun is fired at a 66 ° with a speed of 4.47 m/s.

Find how long the bullet is in the air, how high,

and how far it travels.

“Shooting at an Angle”

velocity

Vv

θ

Vh

First, find the vertical and horizontal components

Vertical Sin 66 ° = Vv Horizontal Cos 66 ° = Vh

4.47 m/s 4.47 m/s

Vv = 4.08 m/s Vh = 1.82 m/s

“Shooting at an Angle”

velocity

4.08 m/s

θ

1.82 m/s

Next, you need to fine the time in the air.

tup = Vv = 4.08 m/s = 0.416 sec

g 9.8 m/s2

ttotal = .416 x 2 = 0 .833 sec.

“Shooting at an Angle”

velocity

4.08 m/s

θ

1.82 m/s

Once the time is known, you can calculate the range and the height

Range = Vh x ttotal = (1.82 m/sec.)(0.833 sec) = 1.52 m 2

“Shooting at an Angle”

velocity

4.08 m/s

θ

1.82 m/s

Once the time is known, you can calculate the range and the height

Height = Vvtup – ½gtup2

= (4.08 m/s)(.416 sec) – (9.8 m/s2)(0.416 sec)2

2

= .849 m

Simulations