#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

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

#2

42

#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.

It´s easy by the way.

You could just google it, when youre that desperate.

edit:^^ ^above poster is right.

#4

Trig

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

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

#5

I'm too busy.

#6

s=ut+1/2ut^2

thatll help

thatll help

#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.

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.

#8

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

#10

s=ut+1/2ut^2

thatll help

That's only if it's going in a vertical motion.

#11

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

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

#13

I'm too busy.

This is what I planned before I entered this thread. Oh well, you got here first

#14

s=ut+1/2ut^2

thatll help

you don't have an angle algebra in that equation.

#15

That's only if it's going in a vertical motion.

And the equation is wrong

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

And the equation is wrong

Yes, I noticed that... <_< >_>

#17

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.

#18

3.

#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)

Then it's just:

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

(using sin a = opposite/hypotenuse)

#20

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

*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.

#22

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

#23

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.

#24

u=175cos(angle)

v=0

s=650

a=9.8

v^2 = u^2 + 2as

v=0

s=650

a=9.8

v^2 = u^2 + 2as

#25

edit:^^ ^above poster is right.

Haha Damn straight, 42 is the answer to life. Also I'm sigging me being right.

>_>

...

<_<

#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

a) find angle of elevation for which the missile would just reach the cloud

#27

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

#28

working out the x component is the hard bit though

I thought I'd see you in here.

#29

you sure s = 1300? becuase 650 is s in the y direction, and i thought that had no effect on the x movement

#30

u=175cos(angle)

v=0

s=650

a=9.8

v^2 = u^2 + 2as

This is all you need. Honestly.

#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

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