Question

A firefighter mounts the nozzle of his fire hose a distance 36.9 m away from the edge of a burning building so that it sprays from ground level at a 45° angle above the horizontal. After quenching a hotspot at a height of 9.45 m, the firefighter adjusts the nozzle diameter so that the water hits the building at a height of 17.1 m. By what factor was the nozzle diameter changed? Assume that the flow rate of water through the hose stays the same, and treat the water as an ideal fluid.

Answer 1

Time, t = x/u cos(theta)

Vertical distance travelled,

y = u sin(theta)t - 0.5 gt^2

9.45 = u sin45(x/u cos45) - 0.5 gt^2

9.45 = 36.9 - (0.5 x 9.8 x t^2)
t = 2.37 sec

Initial velocity, u = 36.9/cos 45 x 2.37 = 22 m/s

17.1 = 36.9 - 4.9 t^2

Time, t = 2.01 sec

New initial velocity, u' = 36.9/ 2.01 x cos45 = 26 m/s

Factor, r2/r1 = sqrt(22/26) =0.91
                            =91%