How long does the object take to reach the ground

  • Thread starter Johnny Leong
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In summary, the initial velocity of the object dropped is u = 0 m/s in all four parts of the question. Using the equations v = u + at and v2 = u2 + 2as, the time taken for the object to reach the ground and its velocity on landing can be calculated for each part. In part (d), the initial velocity is still u = 0 m/s, but if the helicopter is not just about to descend, more information would be needed to find the initial speed.
  • #1
Johnny Leong
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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.
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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?
 

1. How is the time taken for an object to reach the ground calculated?

The time taken for an object to reach the ground is calculated using the equation: t = √(2h/g), where t is the time, h is the height of the object, and g is the acceleration due to gravity (usually taken as 9.8 m/s^2).

2. Will the time taken for an object to reach the ground be the same for all objects?

No, the time taken for an object to reach the ground depends on the height from which it is dropped and the acceleration due to gravity. Objects with different heights or in different gravitational fields (such as on different planets) will have different times to reach the ground.

3. How does air resistance affect the time taken for an object to reach the ground?

Air resistance can affect the time taken for an object to reach the ground by slowing down its acceleration. This means that it will take longer for the object to reach the ground compared to if there were no air resistance. However, for objects with relatively small surface areas (such as a ball or a book), the effect of air resistance is usually negligible and can be ignored in calculations.

4. Can the time taken for an object to reach the ground be changed?

Yes, the time taken for an object to reach the ground can be changed by altering the height from which it is dropped or by changing the gravitational field (such as by dropping the object on a different planet). It can also be changed by introducing external forces such as air resistance or by throwing the object with a certain initial velocity.

5. Is the time taken for an object to reach the ground affected by its mass?

The time taken for an object to reach the ground is not affected by its mass. This is because the equation for time does not involve the mass of the object. However, the acceleration due to gravity does depend on the mass of the object, so objects with different masses will have different accelerations and therefore different times to reach the ground.

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