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Ejercicio 29, página 91

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Transcript of Ejercicio 29, página 91

PROBLEMA
Un jugador patea una pelota con una velocidad que forma un ángulo con la horizontal. Si la pelota lleva una velocidad horizontal de 2m/s y cae a 16m de donde fue lanzada ¿Cuál es la componente vertical de la velocidad de lanzamiento?
DATOS DEL PROBLEMA
-1/2g * t + Y = VoY
---
t



-1/
2
g

*
8
s + 0 = VoY
----
8s
-g * 4s = VoY

- ( -9,8 m/s2) * 4s = VoY

9,8 m/s
2
* 4
s
= VoY

39,2 m/s = VoY
X= Vx*t


Y=VoY * t + 1/2 g * t2
Formulas a aplicar
Ejercicio 29, página 91
Paula Bonilla- Laura Aparicio
X= Vx*t

X/Vx= t

16m/2m/s = 8 s


8s= t
Desarrollo
Formula 2
Y= VoY * + 1/2g * t2

-1/2g * t2 + Y = VoY * t

-1/2g * t2 + Y = VoY
------------------
t

-1/2g *
t2
+ Y = VoY
------------- ---
t
t

-1/2g * t + y = VoY
---
t


Resultado
V0x= 2 m/s
x= 16m
VoY = 32,9 m/s
Full transcript