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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.

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Projectile Motion

• 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

Vh only

Vh only

Vh only

Vh

Vv

• 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

Vh only

h

range

To find the time in the air

use the following equation:

h = gt2

2

Vh only

h

range

Vh only

To find the range, use the time in the air

and the following equation:

range = (Vh)(t)

h

range

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

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

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

• 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.

• 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

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.

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

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.

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

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