A firefighter mounts the nozzle of his fire hose a distance 35.3 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 8.85 m, the firefighter adjusts the nozzle diameter so that the water hits the building at a height of 18.3 m. By what factor was the nozzle diameter changed? Assume that the diameter of the hose stays the same, and treat the water as an ideal fluid.

I tried using .5(4.9)t^2+vosin45t=8.85 and 18.3 then solving for t by plugging in 35.3/cos45 but it gave me 49.7m/s and 49.8

Question

A firefighter mounts the nozzle of his fire hose a distance 35.3 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 8.85 m, the firefighter adjusts the nozzle diameter so that the water hits the building at a height of 18.3 m. By what factor was the nozzle diameter changed? Assume that the diameter of the hose stays the same, and treat the water as an ideal fluid.

I tried using .5(4.9)t^2+vosin45t=8.85 and 18.3 then solving for t by plugging in 35.3/cos45 but it gave me 49.7m/s and 49.8

anonymous · Answer

A1v1=A2v2 is another equation that is useful