A fire hose ejects a stream of water at an angle of [tex]36.3^\circ[/tex] above the horizontal. The water leaves the nozzle with a speed of [tex]21.9 \, \text{m/s}[/tex]. Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?

Using constant acceleration (acc'n) equations: i =horizontal componant, j =vertical (displacement) S={Uit+0.5ait²} + {Ujt+0.5ajt²} (Initial Velocity) U={21.9 cos(36.3)i} + {21.9 sin(36.3)j} (Final Velocity) V={Ui+ait} + {Uj+ajt) (Acc'n) a={0i} + {-9.81j} (time) t=t
When the water is at it's highest point, Vj=0, since no vertical motion is occurring therefore: 21.9 sin(36.3) + -9.81 = 0 therefore, when the water is at it's highest point, t=1.32 (3 sig figures) To find how far away you need to be, you need to work out how far the water would travel in this time, using Si Therefore: when t=1.32..... Si= {21.9 sin(36.3)} (1.32....) + 0.5 9.81*1.32.... = 25.66

Answered by edmc96 | 2024-06-10

To strike the highest possible fire, the fire hose should be located approximately 46.92 meters from the building. This distance is calculated by considering the horizontal component of the water's velocity and the total time of flight. The calculations involve breaking down the initial speed and using the formula for projectile motion.
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Answered by edmc96 | 2024-10-01