A fireman holding a nozzle at a horizontal distance of 4.8 m from a vertical wall wishes to send a jet of water through a small window in the wall located 3.6 m vertically above the nozzle. If the inclination of the jet with the horizontal is 60 degree at the nozzle, calculate the required velocity of the jet at the nozzle exit
Answers
Answer:
let
y
=
1
x
+
4
the denominator of y cannot be zero as this would make
y undefined. Equating the denominator to zero and
solving gives the value that x cannot be
solve
x
+
4
=
0
⇒
x
=
−
4
←
excluded value
domain
x
∈
R
,
x
≠
−
4
(
−
∞
,
−
4
)
∪
(
−
4
,
∞
)
←
in interval notation
to find the range, rearrange making x the subject
y
(
x
+
4
)
=
1
x
y
+
4
y
=
1
x
y
=
1
−
4
y
x
=
1
−
4
y
y
y
=
0
←
excluded value
range
y
∈
R
,
y
≠
0
(
−
∞
,
0
)
∪
(
0
,
∞
)
graph{1/(x+4) [-10, 10, -5, 5]}
xy
Answer:
31.6
m
s
−
1
This solution assumes an incompressible fluid, so the volumetric flow rate is constant at all points in the hose and nozzle (this is because the density is constant and the mass flow rate must be the same at all points).
In order to calculate the of velocity of the water in the nozzle we need to know the cross-sectional area of the nozzle.
A
N
=
π
D
2
N
4
=
π
(
0.022
)
2
4
=
3.8
⋅
10
−
4
m
2
Now we can use the cross-sectional area with the volumetric flow rate to calculate the velocity of the water in the nozzle:
d
V
d
t
=
A
d
s
d
t
Where
d
V
d
t
is the volumetric flow rate and
d
s
d
t
is the rate of change of displacement, i.e. the velocity.
v
=
(
d
V
)
/
(
d
t
)
A
=
0.012
3.8
⋅
10
−
4
=
31.6
m
s
−
1
That is just over 70 mph!