Question

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?

Answer 1

The speed of the mailbag after 5.00 s is 11.6 m/s, and it is 28.9 m below the helicopter. These values remain the same even if the helicopter is rising steadily at 2.32 m/s.

Step-by-step explanation:

(a) To find the speed of the mailbag after 5.00 s, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the mailbag is released from rest, so the initial velocity is 0 m/s. The acceleration is the same as the rising speed of the helicopter, which is 2.32 m/s2. Therefore, the final velocity after 5.00 s is v = 0 + (2.32 m/s2)(5.00 s) = 11.6 m/s.

(b) To find how far the mailbag is below the helicopter, we can use the equation s = ut + (1/2)at2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. In this case, since the mailbag is released from rest, the initial velocity is 0 m/s. The acceleration is the same as the rising speed of the helicopter, which is 2.32 m/s2. Therefore, the distance below the helicopter after 5.00 s is s = (1/2)(2.32 m/s2)(5.00 s)2 = 28.9 m.

(c) If the helicopter is rising steadily at 2.32 m/s, the answers to parts (a) and (b) remain the same as in the previous case. The mailbag would still have a final speed of 11.6 m/s after 5.00 s, and it would still be 28.9 m below the helicopter.

Answer 2

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 +
`(1)/(2)` gt²

where u is 0;

h =
`(1)/(2)` x 9.8 x 5² = 122.5m