Answer:
The rock takes 1.22 seconds to hit the ground
Explanation:
Free Fall Motion
A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.
If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is g = 9.8 m/s^2.
The final velocity of a free-falling object after a time t is given by:
vf=g.t
The distance traveled by a dropped object is:
[tex]\displaystyle y=\frac{gt^2}{2}[/tex]
If we know the height h from which the object was dropped, we can find the time it takes fo hit the ground:
[tex]\displaystyle t=\sqrt{\frac{2y}{g}}[/tex]
[tex]\displaystyle t=\sqrt{\frac{2\cdot 6}{9.8}}[/tex]
[tex]t=1.22\ sec.[/tex]
The rock takes 1.22 seconds to hit the ground
The time taken for the rock to hit the ground from the roof top is 1.1 seconds.
Given the data in the question;
Since the rock was initially at rest
Time taken for the rock to hit the ground; [tex]t = \ ?[/tex]
To determine the time taken for the rock to hit the ground, we use the second equation of motion:
[tex]s = u + \frac{1}{2}at^2[/tex]
Where s is the height or distance, u is the initial velocity, t is time taken and a is acceleration due to gravity( since the rock will be under gravity, [tex]a = g = 9.8m/s^2[/tex] )
We substitute our values into the equation
[tex]6m = 0 + [ \frac{1}{2}\ *\ 9.8m/s^2\ * t^2 ]\\\\6m = 4.9m/s^2 \ * t^2\\\\t^2 = \frac{6m}{4.9m/s^2}\\\\t = \sqrt{\frac{6m}{4.9m/s^2}} \\\\t = 1.1s[/tex]
Therefore, time taken for the rock to hit the ground from the roof top is 1.1 seconds.
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