We have the equation of motion [tex]s = ut + \frac{1}{2} at^2[/tex], where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.
Here displacement = 135 m, Initial velocity = 0 m/s, acceleration = 9.81 [tex]m/s^2[/tex]
Substituting
[tex]135 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2 =135\\ \\ t =5.25 seconds[/tex]
A box falls out of a stationary helicopter hovering 135 m above the ground will take 5.25 seconds to reach the ground.
