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## Motion of a Freely Falling Body

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**Objectives**Explain Galileo’s theory of motion and Newton’s laws of motion. Apply all the equation for motion of an object in free fall. To solve problems on free falling bodies and the acceleration of gravity.**Galileo**The remarkable observation that all free falling objects fall at the same rate was first proposed by Galileo, nearly 400 years ago. • Galileo conducted experiments using a ball on an inclined plane to determine the relationship between the time and distance traveled.**Galileo**He found that the distance depended on the square of the time and that the velocity increased as the ball moved down the incline. • The relationship was the same regardless of the mass of the ball used in the experiment.**The story that Galileo demonstrated his findings by dropping**two cannon balls off the Leaning Tower of Pisa is just a legend. However, if the experiment had been attempted, he would have observed that one ball hit before the other! Galileo**Introduction to Free Fall**A free-falling object is an object which is falling under the sole influence of gravity. That is to say that any object which is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." This definition of free fall leads to two important characteristics about a free-falling object: Free-falling objects do not encounter air resistance. All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s or 32ft/s/s**Freely Falling Body**The acceleration of freely falling body is so important that physicist called it acceleration due to gravity. Denoted by letter gwhich is equivalent to 32ft/sec2 or 9.8m/sec2. Meaning in the 1st second, a falling body accelerates from a stationary position to a velocity of9.8m/sec2, after 2 seconds, the velocity is doubled to 19.6m/sec2 after 3 seconds it triples to 29.4m/sec2**Since accelerating objects are constantly changing their**velocity, you can say that the distance traveled divided by the time taken to travel that distance is not a constant value. A falling object for instance usually accelerates as it falls. The fact that the distance which the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward Freely Falling Body**A simple rule to bear in mind**is that all objects (regardless of their mass) experience the same acceleration when in a state of free fall. When the only force is gravity, the acceleration is the same value for all objects. On Earth, this acceleration value is 9.8 m/s/s that it is given a special name - the acceleration of gravity - and a special symbol - g.**Newton**Newton's first law - the law of interaction states that for every action there's an equal and opposite reaction.**Newton**Newton's second law - the law of acceleration states that the acceleration of an object is directly related to the net force and inversely related to its mass.**Newton’s Law of Acceleration**Fnet = m * a A=F/m**Acceleration depends upon two factors: force and mass.**The 10-kg elephant obviously has more mass (or inertia). This increased mass has an inverse effect upon the elephant's acceleration. And thus, the direct effect of greater force on the 10-kg elephant is offset by the inverse effect of the greater mass of the 10-kg elephant; and so each object accelerates at the same rate - approximately 10 m/s/s.**One Newton is defined as the amount of force required to**give a 1-kg mass an acceleration of 1 m/s/s.**Complete the table**5 m/s/s 10 m/s/s 5 m/s/s 10 1**Free Fall and the Acceleration of Gravity**Free-falling objects are in a state of acceleration. The velocity of a free-falling object is changing by approximately 10 m/s every second. If dropped from a position of rest, the object will be traveling approximately 10 m/s at the end of the first second, approximately 20 m/s at the end of the second second, approximately 30 m/s at the end of the third second, etc.**vf = g t**where g is the acceleration of gravity. An approximate value for g on Earth is 10 m/s/s; more exactly, its value is 9.8 m/s/s. Free Fall and the Acceleration of Gravity • The velocity of a free-falling object which has been dropped from a position of rest is dependent upon the time for which it has fallen. • The formula for determining the velocity of a falling object after a time of t seconds is**Example calculations for velocity**Calculate the velocity of a free-falling object after six, and eight seconds vf = g t Solution At t = 6 s vf = (10 m/s2) (6 s) = 60 m/s At t = 8 s vf = (10 m/s2) (8 s) = 80 m/s**Velocity of FreelyFallingBody**If you were to observe the motion of a free-falling object you would notice that the object averages a velocity of 5 m/s in the first second, 15 m/s in the second second, 25 m/s in the third second, 35 m/s in the fourth second, etc. Our free-falling object would be accelerating at a constant rate.**Distance**The distance which a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of a formula; the distance fallen after a time of t seconds is given by the formula. d = 0.5 g t2 where g is the acceleration of gravity**Calculate the distance fallen by a free-falling object after**one, two and five seconds d = 0.5gt2 Example Calculations: At t = 1 s d = (0.5) (10 m/s2) (1 s)2 = 5 m At t = 2 s d = (0.5) (10 m/s2) (2 s)2 = 20 m At t = 5 s d = (0.5) (10 m/s2) (5 s)2 = 125 m Example calculations for distance**Distance**Given these average velocity values during each consecutive 1-second time interval, the object falls: – 5 meters in the first second,– 15 meters in the second second (for a total distance of 20 meters),– 25 meters in the thirdsecond (for a total distance of 45 meters),– 35 meters in the fourth second (for a total distance of 80 meters).**d=0.5(10g/s2 )(1s)2**v=0.5(10m/s2 )1s d= 20 -5 d=dt -di v=0.5gt d=0.5gt2**The table illustrates that a free-falling object which is**accelerating at a constant rate will cover different distances in each consecutive second. Further analysis of the first and last columns of the table above reveal that there is a square relationship between the total distance traveled and the time of travel for an object starting from rest and moving with a constant acceleration.For objects with a constant acceleration, the distance of travel is directly proportional to the square of the time of travel.**Sample Problem**A coin was dropped from the top of the LTA building with a height of 727 ft. If there is no air resistance, how fast (ft/s) will the coin be moving when it hits the ground? 215.68 ft/s**A marble is dropped from a bridge and strikes the water in 5**seconds. Calculate the speed with which it strikes and the height of the bridge. (Vf = 49 m/s, d = 122.5 m) Problem 1**P2**A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon. T = 1.29 sec**P3**The observation deck of the World Trade Center is 420 m above the street. Determine the time required for a penny to free fall from the deck to the street below. T = 9.26 sec**P4**With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance. V = 94.4 mi/hr**P5**A 10kg block being held at rest above the ground is released. The block begins to fall under only the effect of gravity. At the instant that the block is 2.0 meters above the ground, the speed of the block is 2.5m/sec. The block was initially released at a height of how many meters. D = 2.3 m**Assignment**1. Miguel 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. 2. Brandy 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.**3. A kangaroo is capable of jumping to a height of 2.62 m.**Determine the take-off speed of the kangaroo. 4. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well. 5. Ronald McDonald is riding an Air Balloon on his way to Subic. If Ronald free-falls for 2.6 second, what will be his final velocity and how far will he fall? Assignment**Prepare for an EXAM next meeting….**END OF PRESENTATION**A bowling ball falls freely (near the surface of the Earth)**from rest. How far does it fall in 4 seconds, and how fast will it be going at that time? • The ball's average speed for the first 4 seconds is the average of 0 m/s and 40 m/s, its starting and ending speeds, and distance = average speed times time. • So the object will have fallen 80 meters, and its speed will be 40 m/s.**The position of a free-falling body (neglect air resistance)**under the influence of gravity can be represented by the function where g is the acceleration due to gravity (on earth ) and s0 and v0 are the initial height and velocity of the object (when ).**Example (Falling Object Problem) A ball is thrown vertically**upward from the ground with an initial velocity of 160 ft/s.(a) When will it hit the ground?Solution. Since with and we need This is precisely when or and thus the ball will hit the ground in 10 seconds. with and we need**This illustrates that a free-falling object which is**accelerating at a constant rate will cover different distances in each consecutive second. Further analysis of the first and last columns of the table above reveal that there is a square relationship between the total distance traveled and the time of travel for an object starting from rest and moving with a constant acceleration.For objects with a constant acceleration, the distance of travel is directly proportional to the square of the time of travel.**As such, if an object travels for twice the time, it will**cover four times (22) the distance; the total distance traveled after two seconds is four times the total distance traveled after one second. If an object travels for three times the time, then it will cover nine times (32) the distance; the distance traveled after three seconds is nine times the distance traveled after one second. Finally, if an object travels for four times the time, then it will cover sixteen times (42) the distance; the distance traveled after four seconds is sixteen times the distance traveled after one second.