Bungee question (1 Viewer)

[vice lord]

Yeh for bungee i got him hittin the water, I diffrintiated to chek max displacement bt didnt really bother chekin cause even if he's minimum and hits the water he's going to hit it with max displacement...Did every1 get him hittin the water and does any1 remember the figures?

 

[vice lord]

I got worse....I got like 77m deaded.

Yeh i got somewere about tht i think even a bit more like 79-80m:S

 

[The Nomad]

wait, you had to test the nature of the points

i went on the physical interpretation that when the rope begins to retract, theres a moment where v is zero

Glad to know we did the same thing. Please make sure to tell me what you get on the 16th of December, it'll be interesting to compare marks

Eventually when v = 0, the diver's head reaches 119.5m under the bridge, so he doesn't reach the water...since the water is 125m under. Please confirm. (note that in the equation they gave, when t = 0, x = 82, so their x was taken as the distance from the bridge, not from x = L)

x = e^(-t/10) (29 Sin t - 10 Cos t) + 92.

 

[charlen]

yer i got that he wont hit the water since the most he will go is 119 m

 

[lolokay]

Eventually when v = 0, the diver's head reaches 119.5m under the bridge,

sweet. I think that's what I got (I got 119.x m at least)

 

[Timothy.Siu]

i sorta (tried) to do it by contradiction.

so i subbed in x=123 and did some stuff...hopefully we just get marks for saying NO

 

[cjs4898157]

i got he goes into the water, like

42m under the water LOL

thats what i got first then i thought wtf;; so i re did the calculations by finding the value of t which came out to be like 2 radians or something n subbed in and got less than 120 which should be right.. i hope

 

[-Onlooker-]

To lolokay, untuchablecuz, study-freak and all those other geniuses

Here's an idea.

how about we sub in x = 125 into the eqn

so we get 33e^(t/10) = 29sint - 10cos t

Now, RHS can be changed to 30.67 sin (t - 0.33)

so we have 33e^(t/10) = 30.67sin(t-0.33)

If you can solve that, You got yourself a t-value where x = 125.

however, sketch them and you'd realise the situation is impossible. i.e. x is never 125.

THUS- his head dosen't touch the water

 

[lolokay]

oh lol, there is a really easy way to get the answer without differentiting (as shown in the solutions posted by thuc on here)

 

[-Onlooker-]

oh lol, there is a really easy way to get the answer without differentiting (as shown in the solutions posted by thuc on here)

thuc? where ?

 

[lolokay]

thuc? where ?

http://community.boredofstudies.org/862/mathematics-extension-2/225998/solutions.html

 

[TannouB]

its an easy question:

Differentiate x to get x dot.

Let x dot = 0 for 'max height'

Solve for t (simply divide by e-T or whatever it was)

you should get tan inverse 300/71 or something like that

t = 1 point something seconds.

Substitute into x equation

You get about 114m..

Account for height L leaving 123 - 82 = 41 m, and then YES he does hit the water.

 

[harry4]

so did he or did he not hit the water. there seems to be some conflicting perspectives

 

[The Nomad]

its an easy question:

Differentiate x to get x dot.

Let x dot = 0 for 'max height'

Solve for t (simply divide by e-T or whatever it was)

you should get tan inverse 300/71 or something like that

t = 1 point something seconds.

Substitute into x equation

You get about 114m..

Account for height L leaving 123 - 82 = 41 m, and then YES he does hit the water.

Excuse me? If you substitute t = 0 into the equation they gave, you get x = 82, so obviously the displacement function given in the second part of the question already took into account the fact that the jumper had already travelled 82m. So If he's travelled 114m, plus the 2, that makes 116m, which is less than 125. This means the jumper does not reach the water.

 

[lolokay]

so did he or did he not hit the water. there seems to be some conflicting perspectives

he did not; and this can actually be shown quite simply, without differentiating

 

[TannouB]

Excuse me? If you substitute t = 0 into the equation they gave, you get x = 82, so obviously the displacement function given in the second part of the question already took into account the fact that the jumper had already travelled 82m. So If he's travelled 114m, plus the 2, that makes 116m, which is less than 125. This means the jumper does not reach the water.

lol im so stupid, i didnt think of that, i assumed the +92 was the value for L and that screwed me over... o well (ie i didn't know if they took it into account or not)

Umm yeah your right, which means i got the right answer on that last question ( i said he doesnt hit the water) But woulda lost marks for wrong working somewhere :O

 

[charlen]

lol

 

[-Onlooker-]

I wonder if they'll deduct a mark if we didn't take into account the fact that the person ( mind you his bloody huge) is 2metres tall

 

[annabackwards]

he did not; and this can actually be shown quite simply, without differentiating

Could you explain how?

 

[lolokay]

Could you explain how?

it's here (basically).. http://community.boredofstudies.org/862/mathematics-extension-2/225998/solutions.html

using auxillary method, 29sint - 10sint = sqrt(29^2 + 10^2)sin(t+@) for some value @
and also, e^-t/10 < 1, as t > 0
so x = e^-t/10 *sqrt(29^2 + 10^2)sin(t+@) + 92
< sqrt(29^2 + 10^2) + 92
=~ 122.68

adding the 2 meters of the persons height gives
124.68, which is less than 125 so he does not hit the water