A small mailbag is released from a helicopter that is rising\ steadily at 2.32 m/s.
(a) After 5.00 s, what is the speed of the mailbag?
(b) How far is it below the helicopter?
(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.32 m/s?

Respuesta :

Given parameters:

Velocity of the helicopter  = 2.32m/s

Time given  = 5.00s

Unknown:

a. Speed of the mailbag after 5s

b. How far is the mail bag below the helicopter

Solution:

In this problem, we must apply the appropriate motion equation to solve.

For a;

               v  = u + gt  

 v is the velocity of the mail bag

 u is the initial velocity

 g is the acceleration due to gravity

  t is the time taken

         Notice that the initial velocity of the mail bag is 0;

         V  = 0 + 9.8 x 5 = 49m/s

For b;

  Using;

                    h = ut + [tex]\frac{1}{2}[/tex] gt²

   where u is 0;

                  h  = [tex]\frac{1}{2}[/tex] x 9.8 x 5²  = 122.5m