Water from a garden hose that is pointed 29° 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?

Answer 1

The time that each drop of water remains in the air is
t = (4.7 m)/V cos 29 .
That is based upon how far it travels

The time t must also satissfy
1.4 + (V sin 29) t - (g/2) t^2 = 0
That is based upon the vertical equation of motion.
Substitute t from the first equation into the second equation and solve for V.