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2 52 MECHANICS.
will it go ? How far will it go in a horizontal direction ? How
many seconds will it take to finish the flight?
Solution for height
:
_.9 „:„ 9. _
h =
h =
’2g
1202
X sin.2
55°
64.4
h — 12°2
X °- 81915
"
2
64.4
h
_ 14400 X 0.673
64.4
h = 150.5 feet.
Solving for horizontal range
:
i v2
sin. 2 a
g
Twice the angle of 55° is 110c
and sine of 110° will be sine
of 70°, because 180° —110c = 70° ; therefore, sine of 110° equals
sine of 70° in the second quadrant, and the solution will be :
, _ 1202
X sin. 70°
32.2
14400 X 0.93969
32.2
Solving for time of flight:
v sin. a
128.4 feet.
0.5 £•
120 X sin. 55°
16.1
,_ 120 X 0.81915
1
Y(n
" — 6>1 seconds.
Example.
A nozzle on a hose is placed at an angle of 28° to the
horizontal line and the spouting water when leaving the nozzle
has a volocity of 36 feet per second. How far will it theoretic-
ally reach in a horizontal direction?
Solution:
?/2
sin. 2 a
Range
g
b
_ 362
X sin. 56°
g
, 1296 X 0.82904
O — — 33 37 feet
32.2
—
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