A small mailbag is released from a helicopter that is descending steadily at 1.95 m/s. (a) After 3.00 s, what is the speed of the mailbag?

(b) How far is it below the helicopter?

(a) 31.38 m/s

(b) 38.3 m

Explanation:

(a)

We know that acceleration due to gravity, g, is 9.81 m/.

This means that the descending speed of the mailbag will increase by 9.81m/s every second.

So after 3 seconds its speed will have increased by,

3 x 9.81 m/s = 29.43 m/s

We also already know that the mailbag was descending at a speed of 1.95 m/s before being released from the helicopter.

We can now find the final velocity of the mailbag by adding the change in velocity (29.43 m/s) to the initial velocity (1.95 m/s),

1.95m/s + 29.43 m/s = 31.38 m/s

(b)

We will use the equation , where

s is the distance travelled,

g is the accelaration due to gravity and

t is the time taken to travel the distance.

Using this equation we can determine that in 3 seconds the mailbag has fallen,

In these 3 seconds the helicopter has descended,

3 s x 1.95 m/s = 5.85 m

So the mailbag is now

44.145 m - 5.85 m = 38.3 m (3 s.f.)

below the helicopter.

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures