Nah, Not pi by 4

We have that (i) $x = ut*cos\theta$ and (ii) $y = 160 + ut*sin\theta - 5*t^{2}$ .

From (ii) we find that, when $y = 0$ ,

$t = (\frac{1}{10})*(u*sin\theta + \sqrt{u^{2}*sin^{2}\theta + 3200} \Longrightarrow t = \sqrt{2} (sin\theta + \sqrt{sin^{2}\theta + 16})$ ,

where I used the fact that $u = \sqrt{200} = 10\sqrt{2}$ .

Plugging this result into (i) yields that

$x = 20*cos\theta (sin\theta + \sqrt{sin^{2}\theta + 16})$ .

Taking the derivative of $x$ with respect to $\theta$ yields that

$\dfrac{dx}{d\theta} = (\dfrac{sin\theta}{\sqrt{sin^{2}\theta + 16}} + 1) * cos^{2}\theta - sin\theta(\sqrt{sin^{2}\theta + 16} + sin\theta)$ ,

which, after letting $z = sin\theta$ , can be simplified to

$\dfrac{z + \sqrt{z^{2} + 16}}{\sqrt{z^{2} + 16}} * (1 - z^{2} - z\sqrt{z^{2} - 16})$ .

Now since $z \gt 0$ , for the derivative to equal $0$ we must have that

$1 - z^{2} = z\sqrt{z^{2} + 16} \Longrightarrow (1 - z^{2})^{2} = z^{2}(z^{2} + 16) \Longrightarrow 18 * z^{2} = 1 \Longrightarrow z = \dfrac{\sqrt{2}}{6}$ .

Thus $a = 2, b = 6$ and $a + b = \boxed{8}$ .

Okay, you can't really say that gcd(a,b) = 1, if a = 2 and b = 6, can't you? Better fix this.

There valid with a golden formula year 1993 for this question.

Nice one. Test for application of quadratic equation in reality as well as equation of trajectories nothing else. Simple but maths is required