Disregard the number but please give a detailed response of the answer firefight
ID: 2241829 • Letter: D
Question
Disregard the number but please give a detailed response of the answer
firefighter mounts the nozzle of his fire hose a distance 38.5 m away from the edge of a burning building so that it sprays from ground level at a 45degree angle above the horizontal. After quenching a hotspot at a height of 9.81 m, the firefighter adjusts the nozzle diameter so that the water hits the building at a height of 19.1 m. By what factor was the nozzle diameter changed? Assume that the diameter of the nose stays the same, and treat the water as an ideal fluid.Explanation / Answer
Water aimed upward at 45 degrees and hitting a surface 38.5 meters away at a height of 9.81 meters has an initial speed of v0, let's say, where v0^2 = vx^2 + vy0^2. Then we know
vx = vy0
t = 38.5 m/vx
9.81 m = vy0 t - (1/2) (9.8 m/s^2) t^2
Here we have 3 equations in 3 unknowns.
t = 38.5 m/vy0
9.81 m = (38.5 m/t)(t) - (1/2)(9.8 m/s^2) t^2
(4.9 m/s^2) t^2 = 28.69 m
t = 2.42 seconds
vx = 38.5 m / (2.42 s) = 15.91 m/s
v0 = (15.91 m/s) sqrt(2) = 22.5 m/s
Carrying out the same calculation for 19.1 m instead of 9.81 m, we would have
19.1 m = 38.5 m - (4.9 m/s^2) t^2
t^2 = (19.4 m)/(4.9 m/s^2), and t = 1.98976 s
vx = (38.5 m)/(1.98976 s) = 19.349 m/s
v0 = (19.349 m/s) sqrt(2) = 27.3636 m/s
The nozzle diameter must decrease by a factor (22.5/27.3636)^2 = 0.676,
that is, the diameter should be reduced by 32.4%.