00 s-19.6 m/s² = vf - (+1.50 m/s)vf = -18.1 m/sThe SPEED is therefore 18.1 m/s.Finally.......
(c) We can just repeat the steps for (a) and (b) using +1.50 m/s instead of -1.50 m/s
a = (vf - vo) / t
-9.81 m/s² = (vf - (+1.50 m/s)) / 2.00 s
-19.6 m/s² = vf - (+1.50 m/s)
vf = -18.1 m/s
The SPEED is therefore 18.1 m/s.
Finally...we can work out (b) again using +1.50 m/s instead of -1.50 m/s, but it isn't necessary. Think of it this way. From the helicopter's frame of reference, the bag starts falling from rest in both situations, and would therefore be the same distance below the helicopter at 2.00 seconds in either situation. Calculate it out and you'll see what I mean.
So I hope that helps. Remember to be careful about scalar and vector quantities...they might mess you up if the question asks for one and you provide the other.
I hope that helps. Good luck!
a) Velocity after 2 secs = (gt), = 9.8 x 2, = 19.6m/sec. Add the original down velocity of helicopter, = 21.1m/sec.
b) Distance = (v^2/2g), = (19.6^2/19.6), = 19.6 metres.
a) Velocity after 2 secs. with helicopter ascending at 1.5m/sec.
It will take (v/g) = (1.5/9.8), = 0.1531 secs. for the bag to reach maximum height. So it then drops for (2 - .1531) = 1.8469 secs. Velocity down = (gt), = (9.8 x 1.8469) = 18.1m/sec.
b) Height it falls from maximum height = (v^2/2g), = (18.1^2/19.6), 16.715 metres.
Height achieved from release to maximum height = (v^2/2g), = (1.5^2/19.6), = 0.1148 metres.
(16.715 - 0.1148) = 16.6 metres actual drop in the 2 secs., in which time the helicopter has risen another (1.5 x 2) '= 3 metres. Total = 19.6 metres.
I tend to agree about applying "conventions" to physics. The real world doesn't have + and -, x and y, it is obvious dropping velocity is dropping, it cannot be rising. So why use a -?
"Down" would be more descriptive.
