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WINSTONCH [101]
3 years ago
10

Un bombero alejado d = 31.0 m de un edificio en llamas dirige un chorro de agua desde una manguera contra incendios a nivel del

suelo con un ángulo de θi = 33.0° arriba de la horizontal como se muestra en la figura siguiente. Si la rapidez del chorro cuando sale de la manguera es vi = 40.0 m/s, ¿a qué altura (en m) golpeará el edificio? m
Physics
1 answer:
Mila [183]3 years ago
3 0

Answer:

El chorro golpea el edificio a una altura de 15.943 metros con respecto al suelo.

Explanation:

El chorro de agua exhibe un movimiento parabólico, dado que este tiene una inclinación inicial y la única aceleración es debida a la gravitación terrestre. Las ecuaciones cinemáticas que modelan el fenómeno son:

Distancia horizontal (en metros)

x = x_{o} + v_{o}\cdot t \cdot \cos \theta

Donde:

x_{o} - Posición horizontal inicial, medida en metros.

t - Tiempo, medido en segundos.

v_{o} - Velocidad inicial, medida en metros por segundo.

\theta - Angulo de inclinación del chorro de agua, medido en grados sexagesimales.

Distancia vertical (en metros)

y = y_{o} + v_{o}\cdot t \cdot \sin \theta + \frac{1}{2}\cdot g \cdot t^{2}

Donde:

y_{o} - Posición vertical inicial, medida en metros.

g - Constante gravitacional, medida en metros por segundo al cuadrado.

Partiendo de la primera ecuación, se despeja el tiempo:

t = \frac{x - x_{o}}{v_{o}\cdot \cos \theta}

Si x = 31\,m, x_{o} = 0\,m, v_{o} = 40\,\frac{m}{s} and \theta = 33^{\circ}, entonces:

t = \frac{31\,m-0\,m}{\left(40\,\frac{m}{s} \right)\cdot \cos 33^{\circ}}

t = 0.924\,s

La altura máxima se calcula por sustitución directa de términos en la segunda ecuación. Si y_{o} = 0\,m, v_{o} = 40\,\frac{m}{s}, t = 0.924\,s y g = -9.807\,\frac{m}{s^{2}}, entonces:

y = 0\,m + \left(40\,\frac{m}{s} \right)\cdot (0.924\,s)\cdot \sin 33^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.924\,s)^{2}

y = 15.943\,m

El chorro golpea el edificio a una altura de 15.943 metros con respecto al suelo.

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