The specific question is this:
A target appears at a distance of 1250m with the centre of the target 1.2m above ground level. If the sniper was firing from the prone position, the tip of the barrel 30cm above the ground, what angle would he have to fire at to hit the centre of the target, if the bullet travels at 800m/s?
I tried separating the vector into its components. I found $$1250 = 800\cos(\theta) \times t$$ using the horizontal component of the 800m/s vector, and I found $$ 0.9 = 800\sin(\theta)t - 4.9t^2$$ but solving for theta is difficult here. Using wolfram alpha I come up with $\theta = 0.5893 ^{\circ}$ but no idea how I could solve this in exam conditions, with only a simple calculator. Any ideas?
You can get rid of $\theta$ by using the trigonometric identity: $$\cos^2(\theta)+\sin^2(\theta)=1$$
In this case,
$1250=800t\cos(\theta)$ and $0.9+4.9t^2=800t\sin(\theta)$
$1250^2+(0.9+4.9t^2)^2=800^2t^2(\cos^2(\theta)+\sin^2(\theta))=800^2t^2$
This is a quadratic equation in $t^2$, which should be solvable with a calculator which can take square roots. Upon finding $t$, it is not difficult to get $\theta$.