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ivolga24 [154]
3 years ago
10

Se golpea (chuta) un balón sobre el piso y sale dando botes parabólicos cada vez menores. Si se lanzo inicialmente con una veloc

idad de 32m/s, y un ángulo de 60º y se sabe que en cada bote pierde un cuarto de su velocidad y el ángulo se reduce en 10º, determinar el alcance total logrado al termino del tercer bote y el tiempo empleado en ello Gracias a la persona Desconocida
Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

a)d = 180,91 m

b)t = 11,76 seg

Step-by-step explanation:

Para el lanzamiento de proyectil, la ecuación que nos da la velocidad en V(y) es:

V(y)  = Voy - g*t

en donde Voy = Vo * senα    ( donde Vo es la velocidad inicial, α el angulo del disparo.

Si en esta ecuación hacemos V(y) = 0 estamos en el punto donde el componente en el eje y de la velocidad del proyectil es cero, ese punto es el punto medio del recorrido.

0 =  Vo*sen 60⁰     - g*t

g*t  =  Vo* √3/2

t  = { 32 [m/s] * √3 }2*9,8 [m/s²]

t = 16*√3  / 9,8

t = 2,8278 seg

El tiempo total del primer recorrido es entonces por simetría

t₁ = 2 * 2,8278           t₁  = 5,6556 seg

La distancia del primer impacto al suelo es:

x = Vox * t₁                        ( Vox es constante   Vx = Vo*cos 60⁰ )

x  =  32 * (1/2) * 5,6556

x₁  =  90,49 m

Aplicando los mismos criterios ahora para el segundo bote

Ahora Vo = 32 -  32*(1/4)

V = 24 m/s

g*t  =  24 * sen 50⁰

t =  24* 0,7660/ 9,8  

t =  1,8759

2*t  = 2*1,8759

t₂  = 3,7518 seg

x₂  =  Vox * t₂

x₂  =  24* 0,6428*3,7518

x₂  =  57,88 m

Y para el tercer bote Vo =  24 - 24(1/4)        Vo = 18 m/s     α = 40⁰

t = 18 *0,6428/9,8

t  = 1,18

2t  = t₃  = 2*1,18

t₃ = 2,36 seg

x₃  = Vox * 2,36                Vox = Vo*cos 40      Vox = 18*0,7660  

Vox = 13,79

x₃  = 13,79*2,36

x₃  = 32,54 m

La distancia total será

d = x₁  + x₂ + x₃

d  =  90,49  + 57,88 + 32,54

d = 180,91 m

y el tiempo total será la suma de los tiempos

t =  t₁  +  t₂  +  t₃

t  = 5,65 + 3,75 + 2,36

t = 11,76 seg

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