Respuesta :
We have:
Time (t) = 9.0 seconds
Acceleration (a) = -9.8 m/s²
Final Velocity (v) = 0 m/s ⇒ It's zero as the object's speed stops when it hits the ground
Using one of the constant acceleration equation
s = vt - ¹/₂at²
s = (0×9) - (0.5 × -9.8 × 9)
s = 0 + 44.1
s = 44.1 metres
Time (t) = 9.0 seconds
Acceleration (a) = -9.8 m/s²
Final Velocity (v) = 0 m/s ⇒ It's zero as the object's speed stops when it hits the ground
Using one of the constant acceleration equation
s = vt - ¹/₂at²
s = (0×9) - (0.5 × -9.8 × 9)
s = 0 + 44.1
s = 44.1 metres
Answer:
The distance that it fell is [tex]396.9\rm metre[/tex]
Explanation:
Given information:
Height of the building = [tex]h \rm metre[/tex]
Time=[tex]0.9\rm sec[/tex]
acceleration due to gravity [tex]a=g=9.8\rm m/s^2[/tex]
Taken as positive because the speed is increasing downward
By using, equation of motion,
[tex]h=ut+\frac{1}{2}at^2=0\times t+\frac{1}{2}gt^2[/tex]
[tex]h=\frac{1}{2}9.8\rm m/s^2\times (9s)^2=396.9metre[/tex]
The distance that it fell is [tex]396.9\rm metre[/tex]
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