Answer:
The mailbag will take 2.44 seconds to reach the ground.
Explanation:
The height of a helicopter above the ground is given by:
[tex] h = 3.15\times t^3[/tex]
Height of helicopter at t = 2.10 seconds
[tex]h(2.10 )=3.15\times (2.10 )^3 m=29.17 m[/tex]
The helicopter releases a small mailbag from the height of 29.17 m.
The initial velocity of mailbag = u = 0 m/s
Duration in which mailbag will reach the ground = T
Acceleration due to gravity = g = [tex]9.8 m/s^2[/tex]
Using second equation of motion ;
[tex]s=ut+\frac{1}{2}gt^2[/tex]
We have , s = 29.17
u = 0 m/s
t = T
[tex]29.17m=0 m/s\times T+\frac{9.8 m/s^2\times T^2}{2}[/tex]
Solving for T, we gte :
T = 2.44 seconds
The mailbag will take 2.44 seconds to reach the ground.
