Answer:
The total time to reach ground is 24.89 seconds
Explanation:
Since the height of the helicopter is given by
[tex]h(t)=3.30t^{3}[/tex] thus at time t = 2.30 seconds the height of the helicopter is
[tex]h(2.30)=3.30\times (2.30)^{3}=40.151m[/tex]
The velocity of helicopter upwards at time t = 2.30 is given by
[tex]v=\frac{dh(t)}{dt}\\\\v=\frac{d}{dt}(3.30t^{3})\\\\v(t)=9.90t^{2}\\\\\therefore v(2.30)=9.90\times (2.30)^2=120.45m/s[/tex]
Now the time after which it becomes zero can be obtained using the equations of kinematics as
1) Time taken by the mailbag to reach highest point equals
[tex]v=u+gt\\\\0=120.45-9.81\times t\\\\\therefore t_{1}=\frac{120.45}{9.81}=12.28s[/tex]
2) Time taken by the mailbag to reach ground from a height of 40.151 meters equals
[tex]s=ut+\frac{1}{2}gt^{2}\\\\40.151=120.45t+4.9t^{2}[/tex]
Solving for t we get[tex]t_{2}=0.3289secs[/tex]
Now the total time of the journey is
[tex]\\\\2\times t_{1}+t_{2}\\\\=2\times 12.28+0.3289=24.89secs[/tex]
