Free Fall Definition: The value of g is: Only force acting on the object during its fall is gravity (e.g., No air resistance!) Acceleration of a freely falling object is called: “acceleration due to gravity”, g The value of g is: 9.8 m/s2

Freely Falling Objects In the absence of air resistance ALL objects, regardless of their mass, accelerate towards the Earth at the same rate when dropped from rest acceleration rate due to gravity is given symbol g g =  9.8 m/s2 for all objects close to Earth’s surface

Free Fall-How fast? The velocity of a falling object from rest is given by: v = g  time This means that the velocity of a free falling object increases by 9.8 m/s for each second of the free fall!! Ex: After the 1st second, v = 9.8 m/s After the 2nd second, v = 19.6 m/s After the 3rd second, v = 29.4 m/s Etc…

Free Fall-How far? The displacement, Dd, of a falling object from rest is given by: Dd = ½ g  time2 This means that the displacement of a free falling object increases as the square of the time: Ex, After 1st second, Dd = 1 (½g) = 4.9 m After 2nd second, Dd = 4 (½g) = 19.6m After 3rd second, Dd = 9 (½g) = 44.1m, etc.

Free Fall -- Think About It! During free fall, where is the velocity greatest? What is the acceleration? Is it uniform (constant) or changing? How will two objects of different masses behave when free falling?? Which will hit the ground first? For example, Baseball vs. Feather Book vs. Piece of paper

2 Types of Free Fall Problems Object is launched straight up with an initial upward velocity and returns to initial position Object is dropped from rest and falls straight down (1 – D)

Free Fall – Launched Straight Up t = 3 s vy = 0 stops t = 2 s vy = +9.8 m/s t = 4 s vy = 9.8 m/s t = 5 s vy = 19.6 m/s t = 1 s vy = +19.6 m/s t = 6 s vy = 29.4 m/s t = 0 v0 = +29.4 m/s Time to the maximum height (vy=0) = ½ total time in air

Dropping a Ball Straight Down (Free Fall) A person at the top of a tower DROPS a ball to the ground. It hits bottom in 3 seconds. A. What is its velocity when it hits bottom? B. What was the height from which it was dropped? (Define “UP” as positive) Dd = ½ g  t2 df – di = ½ -9.8 m/s2 ( 9 s2) – di = - 44.1 m di = 44.1 m A. vf = vi + a  t = 0 + -9.8 m/s2 (3 sec) = -29.4 m/s  Define up as positive….X(final) as zero…; g points down, therefore negative 9.8 m/s2 : This makes A “-29.4 m/s”; B “+44.1” m

Free Fall Facts Acceleration rate = g =  9.8 m/s2 going upward and falling downward Ball stops briefly at the maximum height above launch point Flight paths up and down are identical but opposite leaves hand at v = + 10 m/s returns to hand at v =  10 m/s

Throwing a Ball Straight Up Person on the ground THROWS a ball straight up. It hits the ground in 8 seconds. What is its initial velocity? What is its velocity when it returns to your hand? What was its maximum height? (Define UP as positive; Neglect air resistance) t(initial) = t(final) = v(initial) = v(final) = d(initial) = d (final) = Check sign of x… A. vf - vi = g t = - 9.8 m/s2 4 sec vi = 39.2 m/s (From Throw) vf = -39.2 m/s (At ground) Dd= vi  t + ½ g  t2 = (39.2 m/s) (4 s) + ½( - 9.8 m/s2 )(4 s)2 = 78.4 m

Problem Solving Tips Write down “Knowns” = the given values for the variables in the problem Include units and have appropriate units Write down “Unknown” = what the problem is asking Choose appropriate equation and write down equation in algebraic form Substitute in values and solve for unknown quantity

Try the problems given.