time of max water height is ... [20.1 * sin(37.8º)] / 9.8
distance is ... (max time) * [20.1 * sin(37.8º)]
A fire hose ejects a stream of water at an angle of 37.8 ° above the horizontal. The water leaves the nozzle with a speed of 20.1 m/s. 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?
2 answers
Vo = 20.1m/s[37.8o],
Xo = 20.1*Cos37.8 = 15.88 m/s. = Hor. component of Vo.
Yo = Vo*sin37.8 = 20.1*sin37.8 = 12.32 m/s. = Ver. component of Vo.
Y = Yo + g*Tr = 0 at max. ht.
12.32 - 9.8*Tr = 0,
Tr = 1.26 s. = Rise time.
d = Xo*Tr = 15.88 * 1.26 = 20.0 m. from a building.
Xo = 20.1*Cos37.8 = 15.88 m/s. = Hor. component of Vo.
Yo = Vo*sin37.8 = 20.1*sin37.8 = 12.32 m/s. = Ver. component of Vo.
Y = Yo + g*Tr = 0 at max. ht.
12.32 - 9.8*Tr = 0,
Tr = 1.26 s. = Rise time.
d = Xo*Tr = 15.88 * 1.26 = 20.0 m. from a building.
2 answers