Helicopter Problem: Find Time for Mailbag to Reach Ground

  • Thread starter jaytm2291
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In summary: So you would solve for t by differentiating h with respect to t- the initial speed. Do you know how to solve a quadratic equation? That is all you have to do for now.Yes. But I don't even know if these numbers or direction I am going in is correct and you are just telling me to go with what I have been doing. Please tell me I am doing the right thing.I believe that you were using the correct method, you just did not include the initial speed of the helicopter initially. So you would solve for t by differentiating h with respect to t- the initial speed.
  • #1
jaytm2291
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Homework Statement


The height of a helicopter above the ground is given by h = 2.60t3, where h is in meters and t is in seconds. At t = 2.10 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?


Homework Equations


Y=ViT - (1/2)gt^2


The Attempt at a Solution


h=(2.60)(2.10)^3
24.0786=0t-(1/2)(9.8)t^2

Thank you.
 
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  • #2
jaytm2291 said:

The Attempt at a Solution


h=(2.60)(2.10)^3
24.0786=0t-(1/2)(9.8)t^2

Thank you.

That is correct, but your '-' should be a '+' since you will want to take down as the positive direction. Otherwise your displacement is negative.
 
  • #3
rock.freak667 said:
That is correct, but your '-' should be a '+' since you will want to take down as the positive direction. Otherwise your displacement is negative.

Then I would have:

24.0786 = 0t + 4.9t^2
t^2 = 4.914
t=2.2s

This is wrong. Any ideas?
 
  • #4
jaytm2291 said:
Then I would have:

24.0786 = 0t + 4.9t^2
t^2 = 4.914
t=2.2s

This is wrong. Any ideas?

Uhm if that is not correct then it means that the helicopter was rising when the bag was dropped, in which you need to get the initial vertical velocity by differentiating 'h' with respect to 't'.
 
  • #5
rock.freak667 said:
Uhm if that is not correct then it means that the helicopter was rising when the bag was dropped, in which you need to get the initial vertical velocity by differentiating 'h' with respect to 't'.

Vi = h'(t) = 7.8t^2 = 7.8(2.10)^2=34.398 m/s ?
 
  • #6
jaytm2291 said:
Vi = h'(t) = 7.8t^2 = 7.8(2.10)^2=34.398 m/s ?

So now try solving 24.0786=34.398t-(1/2)(9.8)t^2 and see if you get the correct answer.
 
  • #7
rock.freak667 said:
So now try solving 24.0786=34.398t-(1/2)(9.8)t^2 and see if you get the correct answer.

It is a quadratic Equation which gives me..x = 0 when y = .785
 
  • #8
jaytm2291 said:
It is a quadratic Equation which gives me..x = 0 when y = .785

:confused:where did you get x and y from? You only had 't' to solve for.
 
  • #9
Listen I have been trying to figure this problem out for 3 hours and 14 minutes. I'm freaking out over this. Can you just point me in the right direction and tell me which formula to use so I can can get on with this? I still need to do so many other problems and this is really annoying if I'm doing this in the wrong direction.
 
  • #10
jaytm2291 said:
Listen I have been trying to figure this problem out for 3 hours and 14 minutes. I'm freaking out over this. Can you just point me in the right direction and tell me which formula to use so I can can get on with this? I still need to do so many other problems and this is really annoying if I'm doing this in the wrong direction.

What I meant was that you had to solve:

24.0786=34.398t-(1/2)(9.8)t2

to get t, which is why I am confused as to where you got 'x'and 'y' from. I don't mean to cause you anymore frustration but solving for 't' should give the correct answer.
 
  • #11
I got when x is = 0, y is = to .785 is all I know. This is the most frustrating thing I just want to be done with. Please help me finish this problem. Thank you. Can you solve it from the beginning and not take my word for it because I'm really confused. Thank you.
 
  • #12
jaytm2291 said:
I got when x is = 0, y is = to .785 is all I know. This is the most frustrating thing I just want to be done with. Please help me finish this problem. Thank you. Can you solve it from the beginning and not take my word for it because I'm really confused. Thank you.

24.0786=34.398t-(1/2)(9.8)t2

Do you know how to solve a quadratic equation? That is all you have to do for now.
 
  • #13
Yes. But I don't even know if these numbers or direction I am going in is correct and you are just telling me to go with what I have been doing. Please tell me I am doing the right thing.
 
  • #14
I believe that you were using the correct method, you just did not include the initial speed of the helicopter initially.
 

1. How does the helicopter problem relate to real-life scenarios?

The helicopter problem is a common physics problem used to demonstrate the concepts of motion and time. In real-life scenarios, it can be applied to situations involving objects falling from a height, such as a person jumping out of a plane or a ball being dropped from a building.

2. What are the key variables in the helicopter problem?

The key variables in the helicopter problem are the initial height of the helicopter, the acceleration due to gravity, and the final height of the mailbag. These variables are used to calculate the time it takes for the mailbag to reach the ground.

3. How is the time for the mailbag to reach the ground calculated?

The time for the mailbag to reach the ground is calculated using the formula t = √(2h/g), where t is time, h is the initial height of the helicopter, and g is the acceleration due to gravity (9.8 m/s^2). This formula assumes that there is no air resistance present.

4. Can air resistance affect the time for the mailbag to reach the ground?

Yes, air resistance can affect the time for the mailbag to reach the ground. In real-life scenarios, air resistance can slow down the descent of the mailbag and therefore increase the time it takes to reach the ground. In the helicopter problem, air resistance is often ignored for simplicity.

5. How can the helicopter problem be extended to more complex scenarios?

The helicopter problem can be extended to more complex scenarios by incorporating additional variables such as the initial velocity of the helicopter, wind resistance, and the mass of the mailbag. It can also be applied to scenarios involving multiple objects falling at different rates or from different heights.

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