Helicopter and Kinematics Problem

AI Thread Summary
The height of the helicopter is modeled by h=2.90t^3, and the mailbag is released after 1.80 seconds. The velocity of the mailbag at release is equal to the helicopter's velocity at that moment, calculated as v=5.8t^2, resulting in 18.8 m/s. The maximum height reached by the mailbag is confirmed to be 16.9 m, but there is confusion regarding the final velocity and time taken to hit the ground. The discussion emphasizes the importance of understanding kinematic equations and differentiation to solve the problem accurately.
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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