A helicopter is ascending vertically with a speed of 9.00 m/s. At a height of 90 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?
d = Vo*t + 0.5*9.8t^2 = 90 m,
9*t + 0.5*9.8t^2 = 90,
9t + 4.9t^2 - 90 = 0,
4.9t^2 + 9t - 90 = 0,
Solve for t using quad. Formula & get:
t = 3.46 s.
To find the time it takes for the package to reach the ground, we can use the equation of motion for free fall:
h = (1/2)gt^2
Where:
h = height (in this case, 90m)
g = acceleration due to gravity (approx. 9.8 m/s²)
t = time
However, since the package is dropped from a helicopter ascending vertically with a speed of 9.00 m/s, we need to consider that as well.
First, let's find the time it takes for the package to descend from the initial height of 90m to the ground. We can use the equation:
h = ut + (1/2)gt^2
Rearranging the equation to solve for time (t):
(1/2)gt^2 = h
t^2 = 2h/g
t = √(2h/g)
Substituting the known values:
t = √(2 * 90 / 9.8)
t ≈ √(18.37)
t ≈ 4.29 seconds (rounded to two decimal places)
Therefore, it takes approximately 4.29 seconds for the package to reach the ground.