Question

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
A) After 3.00 s, what is the speed of the mailbag ?
B) How far is it below the helicopter ?
C) What are your answers to both A and B if the helicopter was rising steadily at 3.00 m/s ?

Answer 1

Answer:

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

       Speed of mailbag after 3 seconds = 26.4 m/s

       Package is 44.1 meter below helicopter

Explanation:

a)  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

   Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

   v = 3+9.8*3 = 32.4 m/s

  Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$.

    $s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

    Distance traveled by helicopter = 9 meter.

 Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$.

  $s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

  Distance traveled by package  = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

  v = -3+9.8*3 = 26.4 m/s

  Speed of mailbag after 3 seconds = 26.4 m/s

 Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$.

    $s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

    Distance traveled by helicopter = 9 meter.

 Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$.

  $s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

  Distance traveled by package  = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m