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ludmilkaskok [199]

3 years ago

5

Un balón de fútbol que se patea aun angulo de 45° con la horizontal, recorre 30m antes de chocar contra el suelo encuentre, la r

apidez inicial del balón,el tiempo que permanece en el aire, la altura máxima que alcanza

1 answer:

Mamont248 [21]3 years ago

4 0

Responder:

Explicación paso a paso:

El movimiento de la pelota es un movimiento de proyectil ya que se permite que el objeto caiga libremente bajo la gravedad.

Según el rango, el rango del objeto es la distancia cubierta por el objeto en la dirección horizontal.

Rango = U²sin2theta / g

U es la velocidad inicial de la pelota

theta es el ángulo de lanzamiento

g es la aceleración debido a la gravedad

Rango dado = 30 m

Theta = 45 °

Podemos obtener la velocidad inicial del objeto sustituyendo los valores dados en la fórmula

30 = u²sin2 (45 °) /9.81

30 = u²sin90 ° / 9.81

Multiplicación cruzada

294.3 = u²

U = √294.3

U = 17.16 m / s

La velocidad inicial de la pelota es 17.16m / s

- El tiempo que la pelota pasó en el aire es su tiempo de vuelo.

Tiempo de vuelo = 2Usintheta / g

T = 2 (17.16) sen45 ° / 9.81

T = 24.27 / 9.81

T = 2.47 segundos

El tiempo que la pelota pasó en el aire es 2.47 segundos

- La altura máxima de la pelota se expresa como se muestra.

Altura máxima = U²sin²theta / 2g

Altura máxima = 17.16² (sin45 °) ² / 2 (9.81)

Altura máxima = 147.23 / 19.62

La altura máxima alcanzada por la pelota es 7.50m

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