Helicopter and Kinematics Problem

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Homework Help Overview

The problem involves the kinematics of a helicopter and a mailbag it releases. The height of the helicopter is described by a cubic function of time, and participants are tasked with finding the velocity of the mailbag at the moment of release, its maximum height, and its velocity upon hitting the ground.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss whether the velocity of the mailbag at release is zero, with some suggesting it should equal the helicopter's velocity at that time. There are questions about the appropriate time interval for calculating this velocity and considerations regarding the upward acceleration of the helicopter.

Discussion Status

There is an ongoing exploration of the velocity at the moment of release, with some participants providing hints about differentiation and the relationship between displacement and time. Multiple interpretations of the problem are being discussed, particularly regarding the calculations of velocity and height.

Contextual Notes

Participants note discrepancies in calculations and mention that an online service is not accepting certain answers, indicating potential confusion or errors in the mathematical approach being taken.

intriqet
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Homework Statement



The height of a helicopter above the ground is given by h=2.90t^3, where h is in meters and t is in seconds. After 1.80 s, the helicopter releases a small mailbag. Assume the upward direction is positive and the downward direction is negative.

A) what is the velocity of the mailbag when it is released?
B) What max height from the ground does the mailbag reach?
C) What is the velocity of the mailbag when it hits the ground?
D) How long after its release does the mailbag reach the ground?


Homework Equations



Kinematic equations



The Attempt at a Solution



So for the first part shouldn't the velocity at release be zero? if not what time interval would I use to derive velocity at release?

Also, max height should be 2.90(1.80)^3 = 16.9 m

if part a is zero then the final velocity can be derived by equation Vf^2 - Vi^2 = 2a * deltaX.
= -18.2 m/s

the velocity

However, the online service won't accept my answer. :/
 
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would the velocity be zero? the helicopter is moving up, right?

it gives you the formula;

h = 2.6 t^3 which you could also say x = 2.6 t^3

does this look familiar? think about differentiation.

what's the change of displacement with respect to time? (in this case) (\frac{dx}{dt})

from that you can work out it's velocity at release.

just a hint but you might also think, maybe the helicopter is accelerating upwards? it's not a very linear relationship is it? :o so try to think about that as well

(hint: \frac{d^2 x}{dt^2})
 
intriqet said:
So for the first part shouldn't the velocity at release be zero? if not what time interval would I use to derive velocity at release?
No, velocity at release would be equal to the velocity of the helicopter at the time of release.
v = \frac{dh}{dt}
 
2.9t^3 dh/dt = 5.8t^2

V-heli at 1.80 sec = 18.8 m/s = Vi-package which Webassign will not accept.

please help and thanks for the quick reply!
 
2.9 * 3 isn't 5.8! =P
 
Oh i knew that...lol sorry I was in a rush to go to class
 

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