Post a New Question


posted by .

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 -

    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 -

    part c is what I am having trouble with. what equation do I use?

  • physics -

    delta Y (distance of bag below helicopter)
    = Yhelicopter - Ybag

    = Yo + 5.1 t - (g/2) t^2 - Yo -5.1t
    = -(g/2)t^2

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question