
#1
Sep1008, 07:00 PM

P: 66

1. The problem statement, all variables and given/known data
The height of a helicopter above the ground is given by h is in meter and t is in seconds. After 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? 2. Relevant equations v_{xf}= v_{xi}+a_{x}*t v_{x,avg}= (v_{xi} +v_{xf})/2 x_{f}=x_{i}+1/2(v_{xi} +v_{xf})*t x_{f}=x_{i}+v_{xi}*t+1/2a*t^{2} v_{xf}^{2}= v_{xi}^{2}+2*a(x_{f}x_{i}) 3. The attempt at a solution What I did was put 2 sec into the height equation, then use the above bolded equation, where i put the constant of gravit as the acceleration and v_{xi} as 0 and x_{i} as h(2) and x_{f} as 0, but the webassign online says i got it wrong and i did it like 5 times. Can someone tell me their thought process and the answer? Thanks! 



#2
Sep1008, 07:23 PM

HW Helper
P: 5,346

Is there something you left out of the problem statement? The time is given directly by h = 1/2*g*t^{2} 



#3
Sep1008, 07:57 PM

P: 66

I put in 2 seconds to find the height at which the helicopter releases the mailbag, and then put in that height as x_{i}. 



#4
Sep1008, 08:39 PM

HW Helper
P: 5,346

Helicopter Problem!Whoa. That's a whole other ocean. It's got a vertical velocity component up as well as being in free fall. What do you figure is the vertical velocity when it is released? 



#5
Sep1008, 09:01 PM

P: 66





#6
Sep1008, 09:14 PM

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P: 5,346

Does the problem say anything other than that it is just released. Are taking derivatives something you are supposed to be able to do in this course? Do you know how to take dy/dt of y = 3 t^{3} ? 



#7
Sep1008, 09:24 PM

P: 66

If I did use dh/dt, where dh/dt= 9t^{2}, then I guess I could use one of the forumlas where I could use v_{f}, but i still am confused. 



#8
Sep1008, 09:35 PM

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P: 5,346

dh/dt = V_{y} = 9t^{2} and at 2 seconds that value is your upward velocity. 



#9
Sep1008, 09:45 PM

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P: 5,346

Figure then the height that it will continue to go up.
V^{2}/(2*a) = Y That gives you it's maximum height (when you add the height it was dropped at). But you need time. So figure from the first calculated height how long it took. Y = 1/2 a* t^{2} That's time to max height after release. Now take the total height and put it in the same equation again and you have the time to fall. Time up + time down = total time. Voilą. 



#10
Sep1008, 09:46 PM

P: 66





#11
Sep1008, 09:58 PM

P: 66

Okay i got it. Thanks.




#12
Sep1008, 10:00 PM

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P: 5,346




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