Answer:
It takes 2.93 s for the object to reach ground from the height of 42 m.
Explanation:
Given:
Displacement of the object is, [tex]d=-42\ m[/tex]
As it is dropped, initial velocity of the object is, [tex]u=0[/tex]
Acceleration of the object is equal to acceleration due to gravity and is equal to, [tex]a=g=-9.8\ m/s^2[/tex]
Now, using Newton's equation of motion and plugging in the values, we get:
[tex]d=ut+\frac{1}{2}at^2\\-42=0+\frac{1}{2}(-9.8)t^2\\-42=-4.9t^2\\t^2=\frac{-42}{-4.9}\\t^2=8.571\\t=\sqrt{8.571}=2.93\ s[/tex]
So, it takes 2.93 s for the object to reach ground from the height of 42 m.
