Question

A small mailbag is released from a helicopter that is rising\ steadily at 2.32 m/s.

Answer 1

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 + $\frac{1}{2}$ gt²

   where u is 0;

                  h  = $\frac{1}{2}$ x 9.8 x 5²  = 122.5m