GravityGravity

Free falling bodies undergo a constant acceleration.

This constant downward acceleration is gravity.

Earth’s gravity = (g) = - 9.81 m/s²

Gravity (cont.)Gravity (cont.)

When an object is moving at a downward velocity, it is moving in the direction of gravitational acceleration.

1 TON1 TON

(-) gCRASH

(-) v

More GravityMore Gravity

When an object is moving at an upward velocity, it is moving in the opposite direction of gravitational acceleration.(-) g

(+) v

QuestionQuestion

What will fall faster when released at the same height and at the same time…………a text

book or a pencil???

Even More Gravity Even More Gravity The rate at which an object falls is

independent of its mass.

Sample Problem #1Sample 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.

a. What is the velocity of the camera when it hits the ground?

b.How long does it take for the camera to hit the ground?

a. Find vf !

Δ y = - 239 m

a = -3.70 m/s² vi = 0.00 m/s

vf² = vi² + 2aΔy

vf² = 2aΔy

vf = √2aΔy

vf = √2(-3.70 m/s²)(-239 m)

vf = 42.1 m/s down

b. Find Δt !

Vf = Vi + a(Δt)

Δt = Vf / a

vvff = - 42.1 m/s = - 42.1 m/s

a = -3.70 m/s² vi = 0.00 m/s

Δ y = - 239 m

Δt = - 42.1 m/s / -3.70 m/s²

Δt = 11.4 s

Sample Problem #2Sample 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?

2.00 m

Vi = 6.00 m/s

Vf = 0.00 m/s

a = -9.81 m/s²

Step 1: Find out how high the ball reaches.

Vf ² = Vi ² + 2aΔy

Δy = (Vf ²- Vi ²) 2a

Δy = -(6.00 m/s)² 2(-9.81 m/s²)

Δy = -36.0 m²/s² -19.6 m/s²Δy= 1.84 m

y tot= 1.84m + 2.00m = 3.84 m

Step 2: Find out the time of the ball traveling up! Vf = 0.00

m/s

Vi = 6.00 m/s

g = -9.81 m/s²

Vf = Vi + a(Δt)

Δt = (Vf – Vi) / a

Δt = – 6.00 m/s / -9.81 m/s²

Δt up = 0.612 s

Step 3: Find time traveling down.

Δy = -3.84 m

g = - 9.81 m/s²

Vi = 0.00 m/sΔy = Vi (Δt) + ½(a)(Δt)²

0.885 s = Δt down

Δt = √(2Δy)/a

Δt = √2( -3.84 m) /-9.81 m/s²

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