Optimizing Angle for Shooting at a Distance with a Sniper Rifle

[taffman123]

hello - i am a first year university student and received an assignment last week - one question i am having very difficulty with is a question about shooting a rifle gun. the question is

If you wanted to shoot an objec on a building that is 40 m high and and 1500 m far with a velocity of V what would be the angle? is there anyone who can get me on teh rigth track?

 

[Hootenanny]

Hi, sorry posted in the other thread. How do you think you should start?

 

[Hootenanny]

What kinematic equations do you know? Think about the horizontal component because now we have 1500m this is the easiest componet to calculate.

 

[taffman123]

what do u mean?

 

[Hootenanny]

You must know some kinematic equations, v =u +at, for example.

 

[taffman123]

i know 5 kinematic equations

but only 4 i can use because i do not know what v2 is

 

[taffman123]

d = v1t + 1/2at^2 = that is the one i am primarily trying to use
and also i am thinking of using d=vt

 

[Hootenanny]

So, what do you know and what do you need to know? Which equation(s) statisfy the criteria?

 

[taffman123]

with having a d = 1500m
v = 854cosx
and an unknown time it seems easy - but for me i got confused with the cosx

would my time = 1.756cosx? or would it = 1.756/cosx?

 

[Hootenanny]

d = vt is the same as d = ut + \frac{1}{2} at^2 without any acceleration. Which is the case horizontally, therefore you can calculate the horizontal flight time as a function of \theta

 

[taffman123]

(y axis)

d = 40m
t=?
a=-9.8m/s
v1= 854sinx

makin my equation

40=854sinxt + -4.9t^2

 

[Hootenanny]

Your time would be t = \frac{1500}{\cos x}

 

[taffman123]

Your time would be t = \frac{1500}{\cos x}

what happened to the v of 854?

 

[Hootenanny]

Sorry I missed the co-efficent off. My mistake

 

[taffman123]

so when i isolate for t (vertical) would it look like

2t =Square root of (40/854sinx) X 1/-4.9 ? or did i make an isolation mistake

 

[Hootenanny]

I'm afraid you cannot isolate t, the equation is a quadratic.

 

[taffman123]

thats what i tohught because i had a t + t^2 but i just thought if i brought the ^2 over it would be ok

 

[Hootenanny]

No because only one term is squared, when you root both sides you will get t + \sqrt{t} = ....

 

[taffman123]

ok - so now my equation would look like

40/854sinx * 1/-4.9 = t + t^2 and now I am confused

the only thing i could do is substitute t = 1500/854cosx?

 

[taffman123]

but having all these cosx and sinx in the equation is going to fustrate me so much

 

[Hootenanny]

Yes, but I would suggest leaving it in the form 854 \sin x t - 4.6t^2 - 40 = 0

 

[taffman123]

ok so i used the quadratic to solve f or t but in my square root i have sinx now i am totally confused

 

[Hootenanny]

Try substituting in the t = \frac{1500}{854\cos x}into the equation 854 \sin x t - 4.6t^2 - 40 = 0 and using trig identities instead.

 

[taffman123]

i have a question - and hopefully u will not be offended - i am not here for anyone to do it for me - but i was just wondering - have u figured out the answer?

 

[Hootenanny]

No, I'm not offended but I haven't worked it through yet. I'm just telling you what I would do at each stage. I will work it through if you like though.

 

[taffman123]

right now i am at the stage

1.756cosx - 15.1169cos^2x-40= 0

and now i am stuck again

 

[taffman123]

but i did not use a trig idenity yet - so i migth have made a mistake?

 

[Hootenanny]

where's the \sin x gone?

 

[taffman123]

i had

854sinx( 1500/854cosx)

so after multplying i had a sinx on top and bottom so they canceled out but teh sinx was attached to cosx - am i still allowed to cancel it out

 

[Hootenanny]

You've multiplied the fraction wrong

 

[taffman123]

what was my mistake

 

[Hootenanny]

Multiplying fractions:
a \times \frac{b}{c} = \frac{a}{1} \times \frac{b}{c} = \frac{a \cdot b}{c}
So your first term should be:
\frac{1500 \times 854 \cdot \sin x}{854 \cos x} = \frac{1500 \cdot \sin x}{\cos x}

 

[taffman123]

question: where does the t come from - did we not sub in for t?

 

[Hootenanny]

Again, an oversight on my part,must be getting tired.

 

[taffman123]

ok so we have

1500sinx/cosx - 4.9(1500/854cosx)^2 - 40

1500tanx ( correct?) -4.9 (3.085cosx^2) -40 = 0

1500 tanx - 15.1169cos^2x-40 =0

is that correct?

 

[taffman123]

but i still don't see how that helps me find x?

 

[Hootenanny]

Your are correct. You need to get the two trig functions in the same form (sin, cosine, sec, etc.), but at the moment I can't recall any useful trig identities.

 

[taffman123]

i only have learned 2

sinx/cosx = tanx

and sin2x + cos2x = 1

 

[Hootenanny]

It may be worth venturing into the Math forums and asking about the idents.

 

[taffman123]

have u been able to figure out the answer

or would u be able to explain what hadeijv did ? becuase he is finished

 

[Hootenanny]

hadeijv squared the whole equation, giving as topsquark says a bi-quadratic. This allowed himto use the ident \sin^2 x + \cos^2 x = 1 to remove the sin and just leave cosine function.

 

[taffman123]

can i do that in my case as well?

 

[Hootenanny]

Yes, you are both working on the same problem and using the same equations. Looking at the other thread may give you some hints. Topsquark explained it well.

 

[MrRottenTreats]

okay, I am trying to work this out I am at 1500sinx / cosx -15/cos2x - 40 = 0 ... and i do not know where to go form here. and is this correct?