How long does the object take to reach the ground

[Johnny Leong]

I would like to ask the following question about Physics.
Question: An object is dropped from a height of 50 m above ground from a helicopter. How long does the object take to reach the ground and what is its velocity on landing if the helicopter
(a) is at rest,
(b) is ascending with a velocity of 2 m/s,
(c) is descending with a velocity of 2 m/s and
(d) has just ascended to its maximum height and is about to descend at an acceleration 1 m/s2?

My answer:
(a) The initial velocity of the object dropped is u = 0 m/s because the helicopter is at rest. I can use v = u + at and v2 = u2 + 2as to find the answers.
(b) The initial velocity of the object dropped is u = -2 m/s because the helicopter is ascending with a velocity of 2 m/s.
(c) The initial velocity of the object dropped is u = 2 m/s because the helicopter is descending with a velocity of 2 m/s.
The steps to find the answers to (b) and (c) are the same as in (a).
(d) The initial velocity of the object dropped is also u = 0 m/s because the helicopter is just about to descend, no effect on the object. To find the answers, the steps are also the same as in part (a). And the answers to this part should be the same as in part (a) too.

 

[Doc Al]

Just be careful to use a consistent sign convention: for example, I like to use up as positive, down as negative.

Originally posted by Johnny Leong
My answer:
(a) The initial velocity of the object dropped is u = 0 m/s because the helicopter is at rest. I can use v = u + at and v2 = u2 + 2as to find the answers.

Right. Careful about the sign of "a" and "s". I call down negative so both acceleration and distance are negative.

(b) The initial velocity of the object dropped is u = -2 m/s because the helicopter is ascending with a velocity of 2 m/s.

The initial velocity is the same as the velocity of the helicopter. So u = + 2 m/s.

(c) The initial velocity of the object dropped is u = 2 m/s because the helicopter is descending with a velocity of 2 m/s.

If you call down the negative direction, then u = -2 m/s.

(d) The initial velocity of the object dropped is also u = 0 m/s because the helicopter is just about to descend, no effect on the object.

Right.

 

[Johnny Leong]

Thank you for your reply, Doc Al.
I want to raise one more question. If in part (d) of my question, the helicopter is not just about to descend, it has the acceleration 1 m/s2, then how to get the initial velocity of the object dropped and to calculate the answers?

 

[Doc Al]

Originally posted by Johnny Leong
I want to raise one more question. If in part (d) of my question, the helicopter is not just about to descend, it has the acceleration 1 m/s2, then how to get the initial velocity of the object dropped and to calculate the answers?

All you care about is the velocity of the helicopter when the object is dropped. The reason the problem states "just about to descend" is so you can correctly conclude that its vertical speed is zero. (The acceleration is irrelevant.) If it wasn't just about to descend, you would need more information to find the intial speed. Make sense?