Mechanics Study (1 Viewer)

doink

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If you can solve this problem, you are prepared for HSC in Mechanics.

"A Shuttle is launched into the air and is said to reach burnout velocity of 120 m/s before moving. If the acceleration due to drag on it is -g -0.0005v^2 and once reaching its apex it floats to the ground with a constant velocity of 4 m/s find its time of flight" take g=9.81

Ans. 147.7 seconds.

GO

EDIT: Be prepared to use algebra, integration, and frustration driven motivation to still not get the answer.
 

Js^-1

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Oh ffs! I spent an hour on it and about 5 pages, and i got 717.2 seconds. Not even close. Never trust working out done at 3am. By the way, does 'burnout velocity of 120m/s' mean that we can assume it has initial velocity 120m/s? Because that's what I did. I'll try it again tomorow. Awesomely hard question; great post.
 

hon1hon2hon3

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lol suggest i am no where close ? . . . i found my maximun height to be 0.27541. . . . lol . No way . . .But i cant tell what i did wrong .

the equation i got is . a = - (g + 0.0005v^2)
i related using vdv/dx . used the inital condition and found the constant , and used the maximun height v=0 to find the maximun height , no idea where i got it wrong . . ..

Also i also used to find the time . . but i got 12.9 secound . . . . i got something like t= 1/root g tan^-1 root 0.0005 x 120 / root g - 1 / root g tan^-1 root 0.0005 x v / root g . . . . . .

LOL , just sharing . btw . . . please dont tell me the question took upward as negative , or else i got it allllll wrong lol
 

doink

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Yes initial velocity is 120 m/s and the max height is actually 550m.

Answer is actually 149.9 seconds i think.

using vdv=adx
vdv/-g-0.0005v^2 = dx

integrate that to find your x in terms of v and hence your max height when v=0
 

hon1hon2hon3

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For x in terms of v , have u get things like , x = 1/2 ln (( g + 7.2 )/ g + 0.0005v^2)) ? thats what i got
 

doink

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That looks about right, havn't checked the constant fully though.

EDIT: You've made some mistakes with constants, should be 1000ln
 
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