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 the 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 the sinx was attached to cosx - am i still allowed to cancel it out

 

[Hootenanny]

You've multiplied the fraction wrong