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2 years ago

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3. Una máquina lanza un proyectil a una velocidad inicial de 110 m/s , con ángulo de 35°, Calcular: a) Posición del proyectil a

los 6s, b) Velocidad a los 6s, c) Tiempo en la máxima altura, d) Tiempo total del vuelo, e) Alcance logrado.

1 answer:

Semenov [28]2 years ago

5 0

Answer:

a) X = 541,2 m

Y = 199,6 m

b) V(6) = 90,28∠ 2,5⁰

c) t = 6,39 s

d) t(t) = 12,80 s

e) Xmax = 1154,6 m

Explanation: Resolveremos este problema con la aplicación directa de las formulas del movimiento de proyectil

a) Coordenadas en el momento 6 segundos después del disparo

t = 6 s y V₀ = 110 m/s

X = X₀ + V₀*cosα*t Y = Y₀ + V₀*senα*t - (9,8 /2)*t²

Al escoger nuestro sistema de coordenadas en el origen del movimiento

X₀ y Y₀ son 0

Entonces

X = V₀*cosα*t Y = V₀*senα*t - (9,8 /2)*t²

α = 35° cos α = 0,82 sen α = 0,57

X = 110*0,82*6

X = 541,2 m

Y = 110*0,57*6 - 4,9* (6)²

Y = 199,6 m

b) Vₓ es constante a lo largo de todo el movimiento

Vₓ = V₀ₓ = V₀*cos α

Vx = 110 * 0,82 m/s

Vx = 90,2 m/s

Vy = V₀* sen α * - g*t

Vy = 110* 0,57 - 9,8* (6)

Vy = 62,7 - 58,8

Vy = 3,9 m/s

|V| = √ (Vx)² + (Vy)²

|V| = √ (90,2)² + (3,9)²

|V| = √ 8136,04 + 15,21

|V| = 90,28 m/s

Para calcular la dirección de la velocidad

tan θ = Vy/Vx

tanθ = 3,9/90,2

tanθ = 0,043

θ = arctan 0,043

Las tablas indican un valor entre 2 y 3 grados escogeremos 2,5⁰

θ = 2,5⁰

Es interesante notar que la Vy a los 6 segundos es muy baja (cercana a cero) y que también el angulo (2,5⁰) cercano a 0 da idea de que el movimiento está my cerca de la altura máxima (mitad del camino) donde Vy = 0

entonces V(6) = 90,28∠2,5⁰

c) Tiempo para altura máxima

La altura máxima se logra cuando Vy = 0

Vy = V₀* sen α * - g*t Si Vy = 0 entonces

0 = 110* 0,57 - 9,8 *t donde t es el tiempo para altura máxima

t = 6,40 s ( como se había pronosticado )

Entonces el tiempo total del movimiento será 2*(6,40)

t(t) = 12,80 s

e) El alcance logrado será:

Xmax = Vx *t(t)

Xmax = 90,2* 12,80

Xmax = 1154,6 m

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