Question

A small mailbag is released from a helicopter that is descending steadily at 1.08 m/s.(a) After 4.00 s, what is the speed of the mailbag?v = m/s(b) How far is it below the helicopter?d = m(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.08 m/s?v = m/sd = m

Answer 1

Given:

Initial Velocity of the mailbag = 1.08 m/s

Let's solve for the following:

(a). Speed of the mailbag after 4.00s.

To find the speed after 4.00 seconds, apply the motion formula:

v = u + at

Where:

v is the speed after 4.00s

u is the initial velocity = -1.08 m/s (negative because it is descending)

a is the acceleration due to gravity acting on the bag = -9.81 m/s².

t is the time = 4.00 s

Hence, we have:

v = -1.08 + -9.81 x 4.00

v = -1.08 - 39.24

v = -40.32 m/s

Therefore, the speed after 4.00 s is 40.32 m/s.

• (b). , How far is it below the helicopter.

To find the distance between the bag and the helicopter, apply the formula:

$d=ut+\frac{1}{2}gt^2$

Where d is the distance.

u = -1.08 m/s

g = 9.81 m/s²

t = 4.00 s

Plug in values into the equation and solve:

$\begin{gathered} d=-1.08(4.00)-\frac{1}{2}(9.81)(4.00)^2 \\ \\ d=-82.8\text{ m} \end{gathered}$

For the dipslacement of the heliopter, we have:

$d=ut+\frac{1}{2}at$

Where:

a = 0 m/s^2

$\begin{gathered} d=-1.08(4.00)+\frac{1}{2}(0)(4.00) \\ \\ d=-4.32\text{ m} \end{gathered}$

Now, to find the distance, we have:

Distance = (-82.8 m) - (-4.32 m) = -82.8 + 4.32 = 78.48 m

The distance between the bag and the helicopter is 78.48 m

The negative sign is just to indicate the bag is below the helicopter.

• (c.) If the helicopter is risng at 1.08 m/s.

(c.) i) What is the speed of the mailbag?

Apply the fomrula:

v = u - at

Where:

u = 1.08 m/s

a = 9.81 m/s²

t = 4.00 s

Hence, we have:

v = 1.08 - 9.81(4.00)

v = 1.08 - 39.24

v = -38.16 m/s

Therefore, the new speed if the helicoter is rising steadily is 38.16 m/s

(c) ii) The distance.

Apply the fomula:

$d=ut-\frac{1}{2}gt^2$

Where:

u = 1.08 m/s

t = 4.00 s

g = 9.81 m/s^2

Hence, we have:

$\begin{gathered} d=1.08(4.00)-\frac{1}{2}(9.81)(4.00)^2 \\ \\ d=-74.16\text{ m} \end{gathered}$

For the displacement:

$\begin{gathered} d=1.08(4.00)-\frac{1}{2}(0)(4.00)^2 \\ \\ d=4.32\text{ m} \end{gathered}$

Distance = -74.16 m - 4.32 m = -78.48 m

Therfore, the distance between the bag and helicopter if the helicopter is rising steadily is 78.48 m.

ANSWER:

• (,a) 40.32 m/s

,

• (b). 78.48 m

• (c). a) 38.16 m/s

b) d = 78.48 m