Respuesta :
Answer:
kinematic eq. d = Vi*t + 1/2a*[tex]t^{2}[/tex]
Explanation:
Vi=0
d=1/2 a*[tex]t^{2}[/tex]
d= 1/2(9.8)[tex]0.9^{2}[/tex]
d=3.969 m
The distance from which the object falls is 393.9 meters.
Given that, an object falls from a high building and hits the ground in 9.0 seconds and air resistance = 0
So the distance can be calculated by the formula given below.
[tex]s = vt + \dfrac{1}{2} at^2[/tex]
Where, [tex]s[/tex] is the distance in meters, [tex]t[/tex] is time (in seconds) taken by the object in motion, [tex]a[/tex] is the acceleration of the object in meter per second square and [tex]v[/tex] is the final velocity in meter per second.
In this case, object falls from high building to the ground so the final velocity would be zero. Gravitational acceleration is -9.8 meter per second square.
Substituting the values in the above formula,
[tex]s = (9.0\times0) + \dfrac{1}{2}(-9.8\times(9.0)^2)\\s = 0 + (-9.8\times81\times0.5)\\s = 396.9 \;\rm m[/tex]
The distance from which the object falls is 393.9 meters.
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https://brainly.com/question/8898885.
