Free Fall

Review:

Do you remember the four “kinematic equations worth committing to memory” from the last lecture?

Free Fall

For a = constant in time (only!) Then:

Vxf (t) = vxi − at xf (t) = xi + vxi t − ½ at2

Also:

Q1

1 (v xi + v xf ) 2 v xf2 = v xi2 − 2 g ( x f − xi ) vx =

Basic

Derived

Apollo 15 Astronaut David Scott perform Galileo's experiment on the Moon Feather drop

Special Case:

Free Fall

ay=constant= −9.8m/s2

Motion under influence of gravity alone (no air resistance) Near the surface of the earth: a = -g = -9.8 m/s2 Acceleration is always down, regardless of the motion (moving up, moving down, instantaneously at rest) Independent of size, shape, composition, mass

Then:

vyf = vyi − gt yf = yi + vyi t − ½ gt2 Also:

1 (v yi + v yf ) 2 v yf2 = v yi2 − 2 g ( y f − yi ) vy =

Basic

Derived

Q2

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Example: Drop a ball from rest at a height 1.50 m above the ground. (a) How long before it hits the floor? (b) How fast is it moving when it hits?

Kinematics & Calculus Let’s start with a(t) and integrate:

a=

Strategy: 1. Draw a picture. 2. Label the “knowns” and identify the “unknowns”. 3. Choose a coordinate system and positive direction 4. Choose appropriate formulae a. make sure they apply b. try simplest first 5. Work algebra neatly. 6. Practice!

Example: a=constant dv = a dt = a dt

v(t ) = at + C1 Initial condition: vi= v(0) at t = 0

vi = C1 So:

v(t ) = vi + at

dv dt

dv = a dt

dv = a dt

Similarly: Let’s start with v(t) and integrate:

v=

dx dt

dx = v dt

dx = v dt

2

and for a=constant:

v(t ) = vi + at

dx = v dt = (vi + at )dt

x(t ) = vi t + 12 at 2 + C2 Initial condition: xi= x(0) at t = 0

xi = C2 So:

x(t ) = xi + vi t + 12 at 2

Grahically: (a=constant) v(t) vf

vi

lope a =s

atf

½atf2

vitf tf

t

∆x = xf – x i= vitf + ½atf2 = AREA under the curve

Extra Example: Throw a ball up with initial speed vi. A) How high does it rise? B) How long does it take to get to the top? C) How fast is it moving when it comes back to its original height?

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