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AnnZ [28]
3 years ago
7

se lanza un cuerpo desde el origen con velocidad horizontal de 40 m/s, y con un ángulo de 60º. calcular la máxima altura y el al

cance horizontal.
Physics
1 answer:
EastWind [94]3 years ago
7 0

Answer:

1. h = 244.8 m    

2. x = 564.8 m  

Explanation:

1. La altura máxima se puede calcular usando la siguiente ecuación:

v_{f}^{2} = v_{0}^{2} - 2gh     (1)                        

Where:

v_{f_{y}}: es la velocidad final = 0 (en la altura máxima)  

v_{0_{y}}: es la velocidad inicial horizontal en "y"

g: es la gravedad = 9.81 m/s²          

h: es la altura máxima =?

La velocidad incial en "y" se puede calcular de la siguiente manera:

tan(\theta) = \frac{v_{0_{y}}}{v_{0_{x}}}

v_{0_{y}} = tan(60)*40 m/s = 69.3 m/s                    

Resolviendo la ecuación (1) para "h" tenemos:

h = \frac{v_{0_{y}}^{2}}{2g} = \frac{(69.3 m/s)^{2}}{2*9.81 m/s^{2}} = 244.8 m          

2. Para calcular el alcance horizontal podemos usar la ecuación:

x = v_{x}*t

Primero debemos encontrar el tiempo cuando la altura es máxima (v_{f_{y}} = 0).

v_{f_{y}} = v_{0_{y}} - gt    

t = \frac{v_{0_{y}}}{g} = \frac{69.3 m/s}{9.81 m/s^{2}} = 7.06 s      

Ahora, como el tiempo de subida es el mismo que el tiempo de bajada, el tiempo máximo es:

t_{m} = 2*7.06 s = 14.12 s          

Finalmente, el alcance horizontal es:

x = 40 m/s*14.12 s = 564.8 m                                                            

Espero que te sea de utilidad!

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