First, we need to calculate for the height where the mailbag was dropped.
h = 3.50 * t^3
at t = 1.75 s, h is:
h = 3.50 * (1.75)^3
h = 18.758 m
Then we can use the equation to solve for time:
h = v0t + 0.5gt^2
where v0 is initial velocity = 0, t = time, g = acceleration due to gravity
18.758 m = 0 + 0.5 * (9.81 m/s^2) * t^2
t^2 = 3.8242
t = 1.96 s
So the mailbag reaches the ground after 1.96 seconds
To solve the problem we first need to find the height of the helicopter and then use the second equation of motion.
It will take 1.81 seconds for the mailbag to reach the ground.
Given to us
As the function for the height of the helicopter is already given, substituting the value of t,
[tex]h = 3.50 \times t^3\\h = 3.50 \times (1.75)^3\\h = 16.078\ meter[/tex]
Thus, when the mail has dropped the height of the helicopter was 16.078 meters above the ground.
We know the second equation of motion,
[tex]S=ut+\dfrac{1}{2}at^2[/tex]
Substituting the values of the mail,
height of the mail from the ground, S = 16.078 meters,
initial velocity, u = 0,
acceleration due to gravity, a = g = 9.81 m\s²
[tex]16.078=(0)t+\dfrac{1}{2}(9.81)t^2\\\\16.078= \dfrac{1}{2}(9.81)t^2\\\\t^2 = \dfrac{16.078\times 2}{9.81}\\\\t^2 = 3.2778\\\\t= 1.81\ sec[/tex]
Hence, it will take 1.81 seconds for the mailbag to reach the ground.
Learn more about the Equation of motion:
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