FALLING BODIES 



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FALLING BODIES 



of feet it travels; the second is its velocity, or rate of speed; and the 

 third is its acceleration; for a body does not travel at the same rate 

 of speed throughout its fall. The longer it falls, the faster it travels; 

 that is, its velocity increases with every second that it falls. It has 

 been found that the gain in speed of a falling body, its acceleration, 

 in other words, is always the same for each second. The motion of a 

 falling body is described, therefore, as uniformly accelerated motion. 

 This 'means that it is a mathematically exact and unvarying motion, 

 provided the object is acted on by the force of gravity alone. In 

 discussing the laws of falling bodies, v will sometimes be used to repre- 

 sent velocity, a, acceleration, and d, the distance traveled. 



It has been found that the acceleration imparted to a falling body 

 is about thirty-two feet per second; the exact number is 32.16 feet. 

 This never varies, therefore a always stands for 32.16. The velocity 

 of a body at the beginning of the first second of its fall is 0; at the 

 end of the first second, or the beginning of the second second, its 

 velocity is 32.16 feet per second. In the second second, therefore, 

 without taking acceleration into account, a body falls at the rate of 

 32.16 feet per second, the velocity at the beginning of the second. 

 To get the total velocity at the end of the second, the acceleration must 

 be added; therefore the velocity at the end of the second is 64.32 feet 

 per second. Let us, then, state it in this way: the velocity of a falling 

 body at the end of a given second is always the velocity at the begin- 

 ning of that second, plus the acceleration. 



At the end of first second v (0 + 32.16) =32.16 feet per second. 

 At the end of second second v (32.16 + 32.16) =64.32 feet per second. 

 At the end of third second v (64.32 + 32.16) =96.48 feet per second. 

 At the end of fourth second v (96.48 + 32.16) =128.64 feet per second. 



The shorter mathematical formula for finding the velocity of a 

 falling body at the end of any given second, is to multiply 32.16 feet 

 by the number of the second. The result obtained would be the same, 

 of course, as that obtained above by addition. 



The distance traveled in a given second may be found from the 

 average of the velocity at the beginning and the velocity at the end 

 of the second, as follows: 



In the first second d = 



- ie.08 feet 



, ., 32.16 + 64.32 

 In the second second d - ^ - = 48.24 feet 



In the third second d 



In the fourth second d = 



64.32 + 96.48 



= 80.40 feet 



96 ' 48 " 128 ' 64 



feet 



By adding the distance traveled in each separate second, the total 

 distance traveled can be found. In three seconds, therefore, a falling 

 body travels 16.08+48.24+80.40=144.72 feet. Now, 144.72 may also 

 be divided up in this way: 3X3X16.08. The total distance traveled 

 in four seconds, 257.28 feet, may be divided thus: 4X4X16.08. So a 

 shorter mathematical formula has been worked out which states that 

 the distance a falling body travels in a given time may be found by 

 multiplying 16.08 by the square of the number of seconds. This rule 

 you can prove for yourself by the longer formula just outlined. 



The shorter mathematical formula for distance per second is stated 

 in this way: To find the distance a body falls in any given second, 



The illustration shows the length of time required for a body to fall 

 from the top of the building to the ground. From the explanation of the 

 laws of falling bodies, as given in the text, and with the help of the for- 

 mulas in the diagram, the height of the building and the distance the body 

 falls each second may easily be computed by the reader. 



