all you need to know is the speed of a hunk of water headed up.
Q = flow rate = hose area * v
Q = pi (.0159)^2 /4 * 2.4
Q = .000477 m^3/second
but
Q = flow rate = nozzle area * speed
Vi = initial speed up
.000477 m/s = pi r^2 Vi
.000477 = pi (.0064)^2 Vi
Vi = 3.7 m/s
Now we have a plain old throw it straight up problem with initial speed up of 3.7 m/s
v = Vi - 9.81 t
v = 0 at top
t = .373 seconds rising
then
h = 4.9 t^2
h = .68 meters
Water flows at 2.4 m/s through a garden hose of diameter 1.59 cm and emerges from a nozzle of
radius 0.64 cm. If the nozzle is directed vertically upward, to what height would the water rise?
1 answer
2 answers