At what speed does the water leave the nozzle?

Question

Water from a garden hose that is pointed 21° above the horizontal lands directly on a sunbather lying on the ground 4.7 m away in the horizontal direction. If the hose is held 1.4 m above the ground, at what speed does the water leave the nozzle? Please show work.

Answer 1

treat each molecule of water as a point mass.
Then the euqations are
x(t)=v0*cos(th)*t
and
y(t)=1.4-v0*sin(th)*t-.5*9.81*t^2

We know that when y(t)=0, x(t)=4.7, so
4.7=v0*cos(21)*t
and
-1.4=1.4-v0*sin(21)*t-
.5*9.81*t^2
t=4.7/(v0*cos(21))

plug into y(t)
0=tan(21)*4.7-.5*9.8*(4.7/(v0*cos(21))^2

tan21*4.7*2/9.8=4.7^2/(v0^2*cos^2(21))
v0=(4.7/cos(21))/sqrt(tan(21)*4.7*2/9.8)

8.3 m/s
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