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

Answer 1

Let's first calculate velocity of water just after adjustment
We know that this is perfect fluid so the volume in smaller tube after time t must be the same as in higher tube
\[\Delta V= A v \Delta t\]
so
\[A_1v_1\Delta t=A_2v_2 \Delta t\] thus
\[\pi d_1^2/4v_1=\pi d_2^2/4v_2\]
\[\frac{ d_2 }{ d_1 }=\sqrt{\frac{ v_1 }{ v_2 }}=k\] which is our goal( d2 is some fraction of d1. if it's smaller this fraction will be smaller than one and vice versa)It should be smaller.
usin this equation:
\[h=\tan \theta r-\frac{ g }{ 2v^2\cos^2 }r^2\]where r is height and distance from building you should find unknown value which i think is 0,9