Question

A small mailbag is releassed from a helicopter that is ascending steadily at 5m/s. After 2s. a) what is the speed of the mailbag?
b) how far is it below the helicopter?

a) would I use v=v_0 +at
v=5m/s + 9.8m/s^2(2s)
v=24.6 m/s is this the correct speed of the mailbag?
For b) i am certain that i would use h=h_0 + v_0t + 1/2at^2
h=0+ 25(2) +1/2(-9.8)(2)^2
h=50+ -19.6
h=30.4m

are these correct?

Answer 1

You have a sign problem... g is downward, -9.8m/s^2, and vi is positive.

Redo this line..
v=5m/s + 9.8*2

b) The helicopter has went up 10m in 2 seconds, add to that the new position of the bag.
h=0+ 5*2-9.8*2^2 notice t his is negative, so it FELL. Add the magnitudes of the two distances.