Chapter 6 motion in two directions
Download
1 / 8

Chapter 6 Motion in Two Directions - PowerPoint PPT Presentation


  • 58 Views
  • Uploaded on

Chapter 6 Motion in Two Directions. Projectile Motion Projectile- object shot through the air Trajectory- the path that a projectile follows How can horizontal and vertical displacement be determined? Vertical displacement d y = ½ gt 2 + v i t if v i =0 then d y = ½gt 2

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

PowerPoint Slideshow about ' Chapter 6 Motion in Two Directions' - noel-solis


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
Chapter 6 motion in two directions
Chapter 6Motion in Two Directions

Projectile Motion

Projectile- object shot through the air

Trajectory- the path that a projectile follows

How can horizontal and vertical displacement be determined?

  • Vertical displacement

    dy= ½gt2 + vit

    if vi=0 then dy= ½gt2

  • Horizontal displacement

    dx= vxt


Example
Example

  • A stone is thrown horizontally at 15m/s. It is thrown from the top of a cliff 44 m high.

  • How long does it take the stone to reach the bottom of the cliff?

  • How far from the base of the cliff does the stone strike the ground?


Example1
Example

A projectile is launched horizontally from the top of a building with a velocity of 12.7m/s. At what height is the projectile launched if the projectile lands 15.0 m from the side of the building?


Projectiles launched at an angle
Projectiles Launched at an Angle

  • A golf ball is hit and leaves the tee with a velocity of 25.0 m/s at 35.0° with respect to the horizontal. What is the horizontal displacement of the ball?

    V= 25.0 m/s

    = 35.0°

    dx= ?

    t=?


Example2
Example

  • A projectile is fired with a velocity of 19.6 m/s at an angle of 60.0° with the horizontal.

  • Calculate the vertical and horizontal velocity of the projectile.

  • Calculate the time the projectile is in the air.

  • Determine the horizontal displacement that the projectile travels.

  • Determine the maximum height the projectile reaches.


Circular motion
Circular Motion

  • Centripetal acceleration

  • - center-seeking acceleration of an object moving in a circle.

    How is it determined?

    Ac= v2/r or

    Ac= 42r/T2


Centripetal force
Centripetal Force

  • Net force directed toward the center of a circle

    How is it determined?

    Fnet= mac

    Example: A 13g rubber stopper is attached to a 0.93 m string. The stopper is swung in a horizontal circle, making one revolution in 1.18s. Find the tension force exerted by the string on the stopper.


Relative velocity
Relative Velocity

  • Motion is relative to an observer

  • Example: A motor boat traveling 6.0 m/s, East encounters a current traveling 3.8 m/s, South.

  • What is the resultant velocity?

  • If the width of the river is 120 m wide, then how much time does it take the boat to travel shore to shore?

  • What distance downstream does the boat reach the opposite shore?


ad