Describing Motion Free falling …

g = 9.8 m/s2 ACCELERATION DUE TO GRAVITY   All bodies in free fall (that is no air resistance) near the Earth's surface have the same downward acceleration of: g = 9.8 m/s2 NOTE: A body in free fall has the same downward acceleration whether it starts from rest or has an initial velocity in some direction.

gt Okay let’s drop a ball from the top of a cliff… What happens to the velocity each second? gt Vi = 0 m/s Vf =9.8 m/s Vf =19.6 m/s Vf =29.4 m/s Vf = 0 + (9.8 m/s2)(1s) = 9.8 m/s Vf = 9.8m/s + (9.8 m/s2)(1s) = 19.6 m/s or Vf = 0 + (9.8 m/s2)(2s) = 19.6m/s Vf = 0 + (9.8 m/s2)(3s) = 29.4m/s

What happens to the distance traveled each second? 4.9 m 19.6 m 44.1 m 78.4 m 122.5 m d = 0 + ½ (9.8 m/s2)(1s)2 = 4.9 m d = 0 + ½ (9.8 m/s2)(2s)2 = 19.6 m

d (0 m/s)2 + 2(9.8 m/s2)(50m) Vf = 31.3 m/s = (31.3 m/s – 0) Example 1: A ball is dropped from rest at a height of 50 m above the ground. a. What is its speed just before it hits the ground? d vo = 0 m/s d = 50 m g = 9.8 m/s2 vf = ? (0 m/s)2 + 2(9.8 m/s2)(50m) Vf = 31.3 m/s b. How long does it take to reach the ground? = (31.3 m/s – 0) 9.8 m/s2 = 3.19 sec

Let’s look at a coin toss. The coin leaves your hand at 6 m/s and travels upward. The coin speed begins to decrease the moment it leaves your hand. The rate of deceleration is -9.80 m/s2 The velocity is zero at the top of its trajectory.

Then the coin starts the down trip And begins to accelerate downward Vo = 0 at the rate of + 9.80 m/s2 When it gets back to your hand its velocity is? The same velocity that it left your hand with 6 m/s Why?

Let’s look at another example of an up down problem. V = 0 m/s - 9.80 m/s2 down up + 9.80 m/s2

d d d = Vf 2 – Vo2 2a d= 0 + ½ (9.8m/s2)(2.04s)2 d = -(20 m/s)2 Example 2 A stone is thrown straight upward with a speed of 20 m/s. a. How long did the trip take? = 0 – 20 m/s - 9.8 m/s2 vo = 20 m/s vf = 0 m/s g = -9.8 m/s2 = 2.04 s ttotal = 2(tup) = 4.08 s b. How high will the stone go? Or using the down trip… d d d = Vf 2 – Vo2 2a d= 0 + ½ (9.8m/s2)(2.04s)2 d = -(20 m/s)2 2(-9.8 m/s2) = 20.4 m d= 20.4 m

Mrs. See tossed a ball up in the air Mrs. See tossed a ball up in the air. The class observed the ball rise up to a greatest height then fall back into Mrs See’s hand. She then asked the class, “What is the acceleration of the ball at the apex of the motion, which is the highest point reached by the ball?”   Alonzo, Daniel, and Beth said the acceleration was zero. Mary and Frank said it was constant and pointing downward. Cathy and John said it wasn’t zero, but it wasn’t constant either.

Alonzo said the acceleration of the ball was zero because it was momentarily at rest at the highest point of the motion.   Frank said the acceleration was constant and downward because the acceleration of gravity is always constant and gravity was acting on the ball. Beth said the acceleration was zero because it was negative when it was rising, since it was slowing down, and positive when it was falling, since it was speeding up, so it had to pass through zero at the apex of the motion. John agreed with Beth that the acceleration was changing, but it was positive on the way up and negative on the way down because that was the direction the ball was moving. Therefore, it isn’t possible to measure the acceleration of the ball when it is at rest.

Which group do you agree with, or do you have a different opinion?

The Answer   The acceleration of the ball is not zero at the apex of the motion. It is, in fact, downward and constant through the full motion of the ball, just as Frank said.