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Andru [333]
4 years ago
11

Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de

caer de una distancia de 40 m?
Physics
1 answer:
nirvana33 [79]4 years ago
4 0
Hola!

Respuesta:

28,64 m/s.

Explicación:

Datos:

Vo: 6 m/s 
Altura o distancia recorrida: 40 m
Gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula:

Vf²-Vo²=2×g×h.

Es importante hacer notar que al estar lanzando el ladrilo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo se tiene:

Vf
²= (6 m/s)²+ (2)×(9,81 m/s²)x(40m)
Vf²=820,8 m²/s²
√Vf²=√820,8 m²/s² 
Vf= 28,64 m/s

Que tengas un buen día!


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