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Un niño lanza una piedra, desde un acantilado con una velocidad horizontal de 6 m/s. Calcule la distancia “x” y la velocidad con que se estr
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Answer:
x = 6 √ 2y / 9.8
, v = √(2y g)
Explanation:
This is a projectile launching exercise.
There is no acceleration on the horizontal axis
vₓ = x / t
x = vₓ t
on the axis and the initial speed is zero (
= 0)
y = ½ g t²
t = √ 2y / g
t = √2 y / 9.8
this is the time to go down
v = vo – gt
v = 0 – g √ 2y / g
v = √2y g
the horizontal distance is
x = 6 √ 2y / 9.8
in order to complete the calculations you need at the height of the cliff