Gravity and speed?

Question

A small mailbag is released from a helicopter that is descending steadily at 1.00 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 parts (a) and (b) if the helicopter is rising steadily at 1.00 m/s?

Answer 1

As I answered earlier, even though the bag was dropped from the helicopter, the initial velocity is NOT zero because the helicopter was moving. So what is the initial velocity of the bag? If we take down as negative, the helicopter was descending at 1.00 m/s, therefore the bag has an initial velocity of –1.00 m/s. We also take this drop point as our zero position.

A)

KNOWNS

y0 = 0

t = 3.00 s

v0 = -1.00 m/s

UNKNOWN

v

v = v0 – gt

v = -1.00 m/s – (9.80 m/s2)(3.00 s)

v = -30.4 m/s

B)

KNOWNS

y0 = 0

t = 3.00 s

v0 = -1.00 m/s

v = -30.4 m/s

UNKNOWN

y

y = y0 + ½ (v + v0)t = 0 + ½ (-30.4 m/s – 1.00 m/s)(3.00 s)

y = -47.1 m

C)

v = v0 – gt

v = 1.00 m/s – (9.80 m/s2)(3.00 s)

v = -28.4 m/s

y = y0 + ½ (v + v0)t = 0 + ½ (-30.4 m/s + 1.00 m/s)(3.00 s)

y = -44.1 m

Answer 2

gravity is an accelerant force of 9.8 ms-2

velocity after 3 seconds =9.8ms-2 x 3secs=29.4ms-1

distance travelled= 1/2(at^2)=44.1 m

though the helicopter is descending at 1ms-1 for 3 secs

44.1-3.0=41.1m distance between helicopter and mailbag

if the helicopter is ascending then

44.1+3=47.1m