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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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Independence of vectors
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 problems2
“Cliff Problems”

Vh only

Vh

Vv


Cliff problems3
“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 problems4
“Cliff Problems”

Vh only

h

range


Cliff problems5
“Cliff Problems”

To find the time in the air

use the following equation:

h = gt2

2

Vh only

h

range


Cliff problems6
“Cliff Problems”

Vh only

To find the range, use the time in the air

and the following equation:

range = (Vh)(t)

h

range


Cliff problems7
“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 problems8
“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 problems9
“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
“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 angle2
“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 angle3
“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 angle4
“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 angle5
“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 angle6
“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 angle7
“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
Simulations

http://phet.colorado.edu/simulations/sims.php?sim=Projectile_Motion

http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/vectors/mzng.html


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