A fire hose ejects a stream of water at an angle of 37.9 ° above the horizontal. The water leaves the nozzle with a speed of 25.5 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?

Question

i tried 25.5 sin 37.9 but it didn't work

Answer 1

V(ver) = 25.5sin37.9 = 15.66m/s.

The ver. distance is max when velocity
is 0:

Vf = Vo + gt = 0,
15.66 + (-9.8)t = 0,
15,66 - 9.8t = 0,
-9.8d = -15.66,
t(up) = -15.66 / -9.8 = 1.6s.

d(ver) = Vo*t + 0.5gt^2,
d(ver) = 15.66*1.6 + 0.5(-9.8)(1.6)^2,
d(ver) = 25.06 - 10.39 = 14.67m.

d(hor) = 14.67 / tan37.9 = 18.84m = Distance from building.

Answer 2

CORRECTION:

Vo(hor) = 25.5cos37.9 = 20.12m/s.

Vo(ver) = 25.5sin37.9 = 15.66m/s.

The ver distance is max when velocity
is zero:

Vf = Vo + gt = 0,
15.66 + (-9.8)t = 0,
15.66 - 9.8t = 0,
-9.8t = -15.66,
t(up) = -15.66 / -9.8 = 1.6s.

d(hor)=Vo * 2t = 20.12 * 3.2 = 64.3m =
Distance from building.

Answer 3

i ask my teacher he said it wasn't right