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irina1246 [14]
2 years ago
15

Una piedra es lanzada horizontalmente desde la azote de un edificio de 30 metros de altura, con una velocidad de 12 m/s. Hallar

el tiempo que la piedra cae al piso y a qué distancia de la base del edificio llego la piedra.
Chemistry
1 answer:
drek231 [11]2 years ago
3 0

Answer: La piedra tarda 2.47 segundos en caer al suelo, y cae a 29.64 metros de la base del edificio.

Explanation:

Ok, tenemos 2 sistemas de ecuaciones para este problema.

Primero, el vertical:

La aceleración es la aceleración gravitatoria, entonces:

a = -9.8m/s^2

para la velocidad podemos integrar sobre el tiempo y obtenemos

v = (-9.8m/s^2)*t + v0

donde v0 es la velocidad inicial, pero la velocidad inicial es solo horizontal, entonces v0 = 0.

Para la posición integramos de vuelta:

p = (1/2)*(-9.8m/s^2)*t^2 + p0

donde p0 es la posición inicial, en este caso, 30m

p = (-4.9m/s^2)*t^2 + 30m

la piedra va a llegar al piso cuando la posición vertical sea igual a 0m

p = 0m =  (-4.9m/s^2)*t^2 + 30m

de aca podemos despejar el tiempo que la piedra tarda en llegar al suelo.

t = √(30/4.9) segundos = 2.47 s.

Ahora, las ecuaciones para el movimiento horizontal son:

Aceleración nula, pues no hay ninguna fuerza actuando en la dirección horizontal.

a = 0

para la velocidad, integramos sobre el tiempo:

v = 0*t + v0 = v0

donde v0 es la velocidad inicial, en este caso, es 12m/s

v = 12m/s

para la posición integramos de vuelta:

p = 12m/s*t + p0

en este caso podemos asumir que estamos inicialmente en el punto x = x0, asi que la posición inicial es p0 = x0.

p = 12m/s*t + x0

entonces, si queremos calcular la distancia entre la base del edificio y el punto donde cae la piedra, tenemos que calcular:

D = p(2.47s) - p(0s)

D = 12m/s*2.47s + x0 - (12m/s*0s + x0)

D = 29.64m

La piedra tarda 2.47 segundos en caer al suelo, y cae a 29.64 metros de la base del edificio.

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