A helicopter is ascending vertically with a speed of 5.60m/s. At a hight of 115m above the earth, a package is dropped from a window. How much time does it take for the package to reach the ground.
From Vf = Vo - 9.8t, 5.6 = 0 - 9.8t or t = .571sec. to zero verical velocity and maximum height.
From h = Vot - 4.9t^2, h = .56(.571) - 4.9(.571)^2 or h = 1.597 m upward from the 115m drop point for a total height of 116.56m.
Again, from h = Vot + 4.9t^2, the time to reach the ground derives from
116.56 = 0(t) + 4.9t^2 or t = 4.85 sec.
To find the time it takes for the package to reach the ground, we can use the equation of motion for an object in free fall.
The equation to calculate the time taken by an object in free fall is given by:
h = (1/2) * g * t^2
Where:
h is the height from which the object is dropped (115m in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken for the object to reach the ground
Rearranging the equation to solve for t, we get:
t = √(2h/g)
Substituting the given values, we have:
t = √(2 * 115 / 9.8)
t = √(230 / 9.8)
t = √23.47
Therefore, the time it takes for the package to reach the ground is approximately 4.84 seconds (rounded to two decimal places).
Note: This calculation assumes that air resistance is negligible.