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?
i tried 25.5 sin 37.9 but it didn't work
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.
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.
i ask my teacher he said it wasn't right
To solve this problem, you need to find the horizontal distance the water will travel when it reaches its maximum height.
First, let's calculate the initial vertical velocity component of the water. Given that the water leaves the nozzle with a speed of 25.5 m/s and an angle of 37.9° above the horizontal, we can use trigonometric functions to find the vertical component of the velocity.
Vertical velocity component (Vy) = initial speed (v) * sin(angle)
= 25.5 m/s * sin(37.9°)
≈ 25.5 m/s * 0.6018
≈ 15.35 m/s
Next, we'll use this vertical velocity component to find the total time (t) it takes for the water to reach its maximum height. We know that at the maximum height, the vertical velocity component becomes zero (since it stops moving upwards before starting to fall).
Using the kinematic equation:
Vertical displacement (Sy) = Vy * t - 1/2 * g * t^2
0 = 15.35 m/s * t - 1/2 * 9.8 m/s^2 * t^2
This is a quadratic equation in terms of t. Solving it, we get two solutions: t = 0 and t ≈ 3.14 s. Since time cannot be zero in this case, we discard t = 0.
Now, we can find the horizontal distance (Sx) the water travels in this time using its initial horizontal velocity component.
Horizontal velocity component (Vx) = initial speed (v) * cos(angle)
= 25.5 m/s * cos(37.9°)
≈ 25.5 m/s * 0.7984
≈ 20.38 m/s
Horizontal distance (Sx) = Vx * t
= 20.38 m/s * 3.14 s
≈ 64.01 m
Therefore, the fire hose should be located approximately 64.01 meters away from the building to hit the highest possible fire.