1 / 14

Falling Objects Chapter 2.3

Falling Objects Chapter 2.3. Gravity. Free falling bodies undergo a constant acceleration. This constant downward acceleration is gravity . Earth’s gravity = (g) = - 9.81 m/s². 1 TON. 1 TON. (-) v. (-) g. Gravity (cont.).

platt
Download Presentation

Falling Objects Chapter 2.3

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Falling Objects Chapter 2.3

  2. Gravity • Free falling bodies undergo a constant acceleration. • This constant downward acceleration is gravity. Earth’s gravity = (g) = -9.81 m/s²

  3. 1 TON 1 TON (-) v (-) g Gravity (cont.) • When an object is moving at a downward velocity, it is moving in the direction of gravitational acceleration. CRASH

  4. (+) v (-) g More Gravity • When an object is moving at an upward velocity, it is moving in the opposite direction of gravitational acceleration.

  5. Question What will fall faster when released at the same height and at the same time…………a text book or a pencil???

  6. Even More Gravity • The rate at which an object falls is independent of its mass.

  7. Sample Problem #1 A robot probe drops a camera off the rim of a 239 m high cliff on Mars where the free-fall acceleration is -3.70 m/s². • What is the velocity of the camera when it hits the ground? • How long does it take for the camera to hit the ground?

  8. vf = √2(-3.70 m/s²)(-239 m) a. Find vf ! a = -3.70 m/s² vi = 0.00 m/s vf² = vi² + 2aΔy vf² = 2aΔy vf = √2aΔy Δ y = - 239 m vf = 42.1 m/s down

  9. b. Find Δt ! a = -3.70 m/s² vi = 0.00 m/s vf = - 42.1 m/s Δ y = - 239 m Vf = Vi + a(Δt) Δt = Vf / a Δt = - 42.1 m/s / -3.70 m/s² Δt = 11.4 s

  10. Sample Problem #2 Jason hits a volleyball so that it moves with an initial velocity of 6.00 m/s straight upward. If the ball starts from 2.00 m above the floor, how long will it be in the air before it strikes the floor?

  11. 2.00 m Step 1: Find out how high the ball reaches. Vf = 0.00 m/s Vf ² = Vi ² + 2aΔy Δy = (Vf ²- Vi ²) 2a a = -9.81 m/s² Δy = -(6.00 m/s)² 2(-9.81 m/s²) Vi = 6.00 m/s Δy = -36.0 m²/s² -19.6 m/s² Δy= 1.84 m 3.84 m y tot= 1.84m + 2.00m =

  12. Step 2: Find out the time of the ball traveling up! Vf = 0.00 m/s Vf = Vi + a(Δt) Δt = (Vf – Vi) / a g = -9.81 m/s² Δt = – 6.00 m/s / -9.81 m/s² Vi = 6.00 m/s Δt up = 0.612 s

  13. Δy = -3.84 m Step 3: Find time traveling down. Vi = 0.00 m/s Δy = Vi (Δt) + ½(a)(Δt)² g = - 9.81 m/s² Δt = √(2Δy)/a Δt = √2( -3.84 m) /-9.81 m/s² 0.885 s = Δtdown

  14. Step 4: Solve for the total time of ball before it hits the floor. Total time = Δt up + Δt down Total time = 0.612 s + 0.885 s Total time = 1.50 s

More Related