L-4 Free fall & constant acceleration Galileo: In the absence of air resistance, all objects, regardless of their mass, fall to earth with the same acceleration (change in velocity) g g  10 m/s2 = (10m/s)/s  the speed of a falling object increases by 10 m/s every second. This means that if 2 objects start at the same height, they will hit the ground at the same time. This is easier to see on an inclined plane, where the effect of gravity is smaller down the plane. g

Free fall – velocity and distance time (s) speed (m/s) distance (m) d (m) 0.45 4.5 1 10 5 ½10(1)2 2 20 ½10(2)2 3 30 45 ½10(3)2 4 40 80 ½10(4)2 50 125 ½10(5)2 If you drop a ball from the top of a building it gains speed as it falls. Every second, its speed increases by 10 m/s. Also it does not fall equal distances in equal time intervals Galileo recognized this pattern

effect of air resistance: terminal velocity  air resistance increases with speed  m = 100 kg, Fgrav = w = mg = 100 kg 10 m/s2 = 1000 N A person who has their hands and legs outstretched attains a terminal velocity of about 125 mph.

Motion with constant acceleration A ball falling under the influence of gravity is an example of motion with constant acceleration. acceleration is the rate at which the velocity changes with time (increases or decreases) acceleration is the change in meters per sec per second, so it’s measured in m/s2 or ft/s2 or mph/s if we know where the ball starts and how fast it is moving at the beginning we can figure out where the ball will be and how fast it is going at any later time!

Simplest case: constant velocity  acceleration = 0 If the acceleration a = 0, then the velocity is constant. In this case the distance an object will travel in a certain amount of time is given by distance = velocity x time (a=0) d = v  t (for a = 0 only) For example, if you drive at 60 mph for one hour you go 60 mph x 1 hr = 60 mi.

Example – running the 100 m dash Usain Bolt set a new world record in the 100 m dash at 9.58 s! Did he run with constant velocity, or was his motion accelerated? He was not moving in the blocks (at rest), then he began moving when the gun went off, so his motion was clearly accelerated Although his average speed was about 100 m/10 s = 10 m/s, he probably did not maintain this speed all through the race.

running the 100 m dash speed distance start Finish line the winner has the highest average speed = 100 m / time

100 m dash (Seoul 1988)

constant acceleration Example: Starting from rest, a car accelerates up to 50 m/s (112 mph) in 5 sec. Assuming that the acceleration was constant, compute the acceleration. Solution: acceleration (a) = rate of change of velocity with time

The velocity of a falling ball Suppose that at the moment you start watching the ball it has an initial velocity equal to v0 Then its present velocity (v) is related to the initial velocity and acceleration (a) by present velocity = initial velocity + acceleration  time Or in symbols : v = v0 + a  t (for a = constant) [v0 is the velocity when the clock starts (t=0) and v is the velocity at time t later]

Ball dropped from rest If the ball is dropped from rest, that means that its initial velocity is zero, v0 = 0 Then its present velocity = a  t, where a is the acceleration of gravity, which we call g  10 m/s2 or 32 ft/s2, for example: What is the velocity of a ball 5 seconds after it is dropped from rest from the top of the Sears Tower (now the Willis Tower)?  v = 32 ft/s2  5 s = 160 ft/s (109 mph)

The position of a falling ball Suppose we would like to know where a ball would be at a certain time after it was dropped Or, for example, how long would it take a ball to fall to the ground from the top of the Sears Tower (1450 ft). Since the acceleration is constant (g) we can figure this out!

Falling distance or d = ½  g  t2 (g = 10 m/s2) Suppose the ball falls from rest so its initial velocity is zero After a time t the ball will have fallen a distance distance = ½  acceleration  time2 or d = ½  g  t2 (g = 10 m/s2) 5

Falling from the ‘Sears Tower’ After 5 seconds, the ball falling from the Sears Tower will have fallen distance = ½  32 ft/s2  (5 s)2 = 16  25 = 400 feet. We can turn the formula around to figure out how long it would take the ball to fall all the way to the ground (1450 ft)  time = square root of (2 x distance/g)

Look at below! or when it hit the ground it would be moving at v = g  t =32 ft/s2  9.5 sec = 305 ft/s or about 208 mph (watch out!)

How high will it go? v = 0 for an instant Let’s consider the problem of throwing a ball straight up with a speed v. How high will it go? As it goes up, it slows down because gravity is pulling on it. At the very top its speed is zero. It takes the same amount of time to come down as it did to go up. going down: v = vo + gt, where, vo = 0, so v = gt  t = v/g going up: tup = vo/g vo

Example A volleyball player can leap up at 5 m/s. How long is she in the air? SOLUTION total time = ttotal = tup + tdown time to get to top = tup = vo / g where vo is the initial upward velocity tup = 5 m/s / 10 m/s2 = ½ sec ttotal = ½ s + ½ s = 1 s

An amazing thing! When the ball comes back down to ground level it has exactly the same speed as when it was thrown up, but its velocity is reversed. This is an example of the law of conservation of energy. We give the ball some kinetic energy when we toss it up, but it gets it all back on the way down. vo vo

So how high will it go? If the ball is tossed up with a speed v, it will reach a maximum height h given by Notice that if h = 1m, this is the same velocity that a ball will have after falling 1 meter.

Escape from planet earth (Not everything that goes up must come down!) To escape from the gravitational pull of the earth an object must be given a velocity at least as great as the so called escape velocity For earth the escape velocity is 7 mi/sec or 11,000 m/s, 11 kilometers/sec or about 25,000 mph. An object given this velocity (or greater) on the earth’s surface can escape from earth!

Formulas apply whenever there is constant acceleration - example a car moving at v0 = 3 m/s begins accelerating at a = 2 m/s2. When will its velocity increase to 13 m/s? SOLUTION: v = v0 + a  t 13 m/s = 3 m/s + 2 m/s2  t 13 m/s = 3 m/s + 10 m/s  t = 5 seconds

Example – deceleration – slowing down deceleration means that the acceleration is opposite in direction to the velocity Suppose you are moving at 15 m/s and apply the brakes. The brakes provide a constant deceleration of – 5 m/s2. How long will it take the car to stop? v = v0 + a t 0 = 15 m/s + (–5m/s2) t  t = 3 s

Another example To spike the ball, a volleyball player leaps 125 cm straight up. What was her speed when she left the court? formula  125 cm = 1.25 m