EYDROKINETICS. 14? 



(c.) In the solution above we were obliged to find the number of 

 seconds that would be required for a body to fall a distance equal 

 to the head, before we could use the formula for the velocity. It is 

 desirable, if possible, to shorten this circuitous process from two 

 stages to one. This we may do as follows : 



Substituting this value of t in the formula, v = gt, 



But h (the head) = 8. Substituting this value of 8 in the last 

 equation, we have, for the velocity of streams issuing from orifices, 

 the following formula : 



v = y 



The value of g being taken in feet, li and v must represent feet 

 also. 



(d.) With what velocity will water issue from an orifice under a 

 head of 144.72 feet ? 

 = 8.02 ft 



u = 8.02 v / i44T72"= 8.02 x 12.03 = 96,48, the number of feet. 



255. Orifice of Greatest Range. The path of 

 a stream spouting in any other than a vertical direction is 

 the curve called a parabola ( 135). The range of such a 

 stream will be the greatest when it issues from an orifice 

 midway between the surface of the liquid and the level of 

 the place where the stream strikes. Streams flowing from 

 orifices equidistant above and below this orifice of greatest 

 range will have equal ranges. (See Fig. 86.) The range, 

 in any such case, may be calculated by the laws of pro- 

 jectiles. 



. (a.) Given an aperture four feet below the surface and 20 ft. 

 above the point where the water strikes, to find the range of the 

 jet 



<o 8.02 ^/h = 8.02 x 2 = 16.04 ft. per second. 

 8=lfft> 



20 - 16.08* 3 /. t = 1.11 + sec. 

 Range = 16.04 ft. x 1.11 = 17.8044 ft. 



