Motion of heavy Bodies* 171 



we learn that in order to find the space or height s through which 

 a heavy body falls in a number t of seconds, we have only to 

 multiply the square of this number of seconds by ^ g, that is, 

 by the space described in the first second. 



Hence, the height or number of feet through which a heavy body 

 falls during a number t of seconds is so many times 16,1 feet as there 

 are units in the square of this number of seconds. 



Thus, when a body has been suffered to fall freely during 7 

 seconds, we may be assured that it has passed through a space 

 equal to 49 times 16,1 feet, or 788,9 feet. We see, therefore, 

 that when, in the case of falling bodies, the time elapsed is 

 known, nothing is more easy than to determine the velocity ac- 

 quired, and the space described. 



276. If the question were to find the time employed by a 

 body in falling from a known height, the equation s = ?gt 3 9 



gives t 2 = j , and consequently, 



t = 



that is, we seek how many times the height s contains 1 g, or 16,1 

 feet, the space described by a body in the first second of its fall, 

 and take the square root of this number. 



277. If we would know from what height a heavy body must fall 

 to acquire a given velocity, that is, a velocity by which a certain 

 number of feet is uniformly described in a second ; from the 



equation v = g J, I deduce the value of , namely, t = ; sub- 

 stituting this value in the equation s = \gt 2 , I have 



T? a 7>2 



s = 1 g X = , 



by which I learn, that in order to find the height s from which a 

 heavy body must fall to acquire a velocity T, of a certain num- 

 ber of feet in a second, the square of this number of feet is to be 

 divided by double the velocity acquired by a heavy body in 

 one second, that is, by 64,4, 



