The height of a helicopter above the ground is given by h = 3.30 t 3 , where h is in meters and t is in seconds. At t = 2.30 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Question

The height of a helicopter above the ground is given by   h  =  3.30      t    3  , where h is in meters and t is in seconds. At t = 2.30 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Lexi Jordan · Accepted Answer

Since the helicopter&#039;s height is determined by  h  (  t  )  =  3.30      t    3   thus at time t = 2.30 seconds the height of the helicopter is  h  (  2.30  )  =  3.30  ×  (  2.30      )    3    =  40.151  mAt time t = 2.30, the helicopter&#039;s upward velocity is given by  v  =            d      h      (      t      )              d      t          v  =      d          d      t        (  3.30      t    3    )    v  (  t  )  =  9.90      t    2      ∴  v  (  2.30  )  =  9.90  ×  (  2.30      )    2    =  120.45  m      /    sNow the time after which it becomes zero can be obtained using the equations of kinematics as1) Time taken by the mailbag to reach highest point equals  v  =  u  +  g  t  0  =  120.45  −  9.81  ×  t    ∴      t    1    =      120.45    9.81    =  12.28  s2) Time taken by the mailbag to reach ground from a height of 40.151 meters equals  s  =  u  t  +      1    2    g      t    2      40.151  =  120.45  t  +  4.9      t    2  Solving for t we get       t    2    =  0.3289  s  e  c  sThe duration of the trip is now  2  ×      t    1    +      t    2    =  2  ×  12.28  +  0.3289  =  24.89  s  e  c  s

The height of a helicopter above the ground is given by h = 3.30t^3, where h is in meters and t is in seconds. At t = 2.30 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

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