Acceration and Free Fall v

Do heavier objects fall faster than lighter ones when starting from the same position? Does air resistance matter? If the free fall motion has a constant acceleration, what is this acceleration and how was it found? How do we solve problems involving free fall?

 Kinematic Equations  the branch of mechanics concerned with the motion of objects without reference to the forces that cause the motion.  Two types of Equations  Constant Motion Equations (accleration = 0)  Constant Acceleration questions (velocity is constantly changing) (a≠0 but is a constant…the same)

 A position-time graph of a bike moving with constant acceleration is shown on the right. Which statement is correct regarding the displacement of the bike? A. The displacement in equal time interval is constant. B. The displacement in equal time interval progressively increases. C. The displacement in equal time interval progressively decreases. D. The displacement in equal time interval first increases, then after reaching a particular point it decreases.

 Galileo concluded:  neglecting the effect of the air, all objects in free fall had the same acceleration.  It didn’t matter what they were made of, how much they weighed, what height they were dropped from, or whether they were dropped or thrown.  The acceleration of falling objects, given a special symbol, g, is equal to 9.80 m/s 2.  The acceleration due to gravity is the acceleration of an object in free fall.

Free Fall is the motion of the body when air resistance is negligible and the action can be considered due to gravity alone. Free fall Video

Even they have a different weight? Yes Even if they are different sizes and shapes? Yes However, remember its YES as long as we do not include air resistance ( for example a feather falls at the same rate but is very affected by air resistance where the lacrosse ball is affected very little.

A ball is dropped from rest from the top of a building. Find: a) The instantaneous velocity of the ball after 6 sec. b) How the ball fell. c) The average velocity up to that point. Answers: -60m/s, 180m, -30m/s

A hammer is dropped on the moon. It reaches the ground 1s later. If the distance it fell was 0.83m: 1. Calculate the acceleration due to gravity on the surface of the moon. 2. Calculate the velocity with which the hammer reached the ground and compare to the velocity it would have, if it was dropped on the earth’s surface. Answer:-1.66m/s 2, -1.66m/s, -9.8m/s

 Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground.

 An inspection of the four equations above reveals that the equation on the top left contains all four variables.four equations above d = v i t + ½ a t 2  Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.  m = (0 m/s) (t) + ½ (-9.8 m/s 2 ) (t) 2  m = (0 m) *(t) + (-4.9 m/s 2 ) (t) 2  m = (-4.9 m/s 2 ) (t) 2  (-8.52 m)/(-4.9 m/s 2 ) = t 2  s 2 = t 2  t = 1.32 s  The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.)

 The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate.  The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable.  The calculated time easily falls within this range of reasonability.  Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

 Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Diagram:Given:Find: v i = 26.2 m/s v f = 0 m/s a = -9.8 m/s 2 d = ?? v f 2 = v i 2 + 2ad

 (0 m/s) 2 = (26.2 m/s) (-9.8m/s 2 ) d  0 m 2 /s 2 = m 2 /s 2 + (-19.6 m/s 2 ) d  (-19.6 m/s 2 ) d = 0 m 2 /s m 2 /s 2  (-19.6 m/s 2 ) d = m 2 /s 2  d = ( m 2 /s 2 )/ (-19.6 m/s 2 )  d = 35.0 m  The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.)