#1
Wonder if anyone could help me with a quick question.

If a missile is fired from ground level and aimed at angle x, initial velocity of 175m/s, and just reaches a cloud at 650 m above ground level, is there anyway of finding what angle x is?

Much obliged xxx
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#2
42
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#3
do your math homeworks yourself.

It´s easy by the way.

You could just google it, when youre that desperate.


edit:^^ ^above poster is right.
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#4
Trig

Pretty simple if you put the x/y components down into a right angle triangle.
#7
Trig.
You have the hypotenuse (650) and the adjacent (175) so you use cosine.
Cosine(angle)=A/H
(Angle)=Cosine^-1(175/650)

I think that'd be it.
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#8
Quote by Random88
s=ut+1/2ut^2

thatll help


I swear he doesn't need eq' of motion.. its just simple trigonometry

EDIT: ^ Yes
#9
x=(nonsense of this thread / Angle of my Dick + Irrevelant 4chan Postings) *Sinus Alpha= a/Hypotenuse
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Also, I like black.


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#10
Quote by Random88
s=ut+1/2ut^2

thatll help

That's only if it's going in a vertical motion.
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#11
Quote by opc100
Trig.
You have the hypotenuse (650) and the adjacent (175) so you use cosine.
Cosine(angle)=A/H
(Angle)=Cosine^-1(175/650)

I think that'd be it.



You wouldn't measure the height of the cloud diagonally, would you?
#12
Quote by vitchb
Trig

Pretty simple if you put the x/y components down into a right angle triangle.


working out the x component is the hard bit though
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Quote by Aléx
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#15
Quote by opc100
That's only if it's going in a vertical motion.


And the equation is wrong

Quote by stevo_epi_SG_wo
working out the x component is the hard bit though


He doesn't actually need the x component, he's already got 2 bits of info.
But even if he needed X its just Pythagoras
Last edited by vitchb at Oct 11, 2009,
#16
Quote by vitchb
And the equation is wrong

Yes, I noticed that... <_< >_>
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#17
Quote by vitchb

He doesn't actually need the x component, he's already got 2 bits of info.
But even if he needed X its just Pythagoras


Its not Pythagoras, he has a side length and a velocity, not two sides.
#19
Surely you would need to know the length of time the missile has been travelling for?

Then it's just:

angle = sin^-1(650/(time*175))

(using sin a = opposite/hypotenuse)
#20
Quote by vitchb
And the equation is wrong

shhh. im hungover

but that would be the way to do it. the 'u' is just the vertical component of the velocity (175cos(theta)), theta being the angle. Then you put it all into

s=ut + 1/2 at^2

and you know everything except theta, so bingo

EDIT: Just realised im talking sh*te, you dont know t
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Last edited by Random88 at Oct 11, 2009,
#21
there's not enough information... what does a cloud have to do with anything? you can fire something at 175 m/s and it will hit a cloud at alt. 650 with a large range of angles... the only thing i can tell you is that they're going to be toward the low end of the acute range.
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#22
Quote by opc100
Trig.
You have the hypotenuse (650) and the adjacent (175) so you use cosine.
Cosine(angle)=A/H
(Angle)=Cosine^-1(175/650)

I think that'd be it.


I thought that originally, but 175 is not the distance, but the velocity its travelling at. cheers anyway though
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#23
Quote by Bronek
there's not enough information... what does a cloud have to do with anything? you can fire something at 175 m/s and it will hit a cloud at alt. 650 with a large range of angles... the only thing i can tell you is that they're going to be toward the low end of the acute range.


I'm guessing 650 is the maximum height, its just worded in a really bad way.
#25
Quote by Dawnwalker

edit:^^ ^above poster is right.


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#26
The question says a missile is projected from a point O on a horizontal plane with speed 175ms-1. The horizontal lower surface of the cloud is 650m above the ground.

a) find angle of elevation for which the missile would just reach the cloud
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#27
Quote by Lakey
I'm guessing 650 is the maximum height, its just worded in a really bad way.


now we're talking :P

therefore the vertical component of the equation would be calculated from assuming the object will travel 1300 m at an initial speed of 175 m/s with an acceleration of -10 m/s^2.

s=1300
a=-10
t=?
v0= 175

s = v0*t + (a*t^2)/2

1300= 175t -5t^2 figure out the rest
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#28
Quote by stevo_epi_SG_wo
working out the x component is the hard bit though


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#29
you sure s = 1300? becuase 650 is s in the y direction, and i thought that had no effect on the x movement
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#30
Quote by Random88
u=175cos(angle)
v=0
s=650
a=9.8

v^2 = u^2 + 2as

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#31
Ok, so just help me work throught this (im abit slow) Btw wouldn't a be -9.8 as gravity is working against it?

v^2 = u^2 + 2as
0=175 sin pheta^2 + 2 x -9.8 x 650
0=175 sin pheta^2 - 12740
12740 = 175 sin pheta^2
112.87 = 175 sin pheta
0.64497 = sin pheta
40 = pheta??

****ING GET IN THERE, THANKS ALOT RANDOMN
say hello, to my little friend!!

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