A helicopter is rising at 5.1 m/s when a bag of its cargo is dropped. (Assume that the positive direction is upward.)
(a) After 2.0 s, what is the bag's velocity?
(b) How far has the bag fallen?
(c) How far below the helicopter is the bag?
physics - drwls, Monday, September 13, 2010 at 11:20pm
Use the equations
V = 5.1 - g t
Y = Yo + 5.1 t - (g/2) t^2
Yo is the elevation where the release occurs. You know what g is, I'm sure.
For (c), remember that the helicopter will be 10.2 m higher than Yo at t = 2 s, sincve it continues to rise. The bag will be lower than Yo.
physics - penny, Monday, September 13, 2010 at 11:32pm
part c is what I am having trouble with. what equation do I use?
physics - drwls, Monday, September 13, 2010 at 11:36pm
delta Y (distance of bag below helicopter)
= Yhelicopter - Ybag
= Yo + 5.1 t - (g/2) t^2 - Yo -5.1t