gravity free falling bodies undergo a constant acceleration. this constant downward acceleration is...
TRANSCRIPT
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