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.
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?
3 answers
part c is what I am having trouble with. what equation do I use?
delta Y (distance of bag below helicopter)
= Yhelicopter - Ybag
= Yo + 5.1 t - (g/2) t^2 - Yo -5.1t
= -(g/2)t^2
= Yhelicopter - Ybag
= Yo + 5.1 t - (g/2) t^2 - Yo -5.1t
= -(g/2)t^2
3 answers