To solve this problem without calculus, I would need to prove a certain theorem.
Let us think about the following problem.
alt text What is the minimum velocity $\displaystyle u_{min}$ for which the particle passes through $\displaystyle (a,b)$ ?

Considering the trajectory equation for the motion of the particle,

$y = x\tan\theta - \frac{gx^2}{2u^2\cos\theta^2}$

Substituting $\displaystyle (a,b)$ for $\displaystyle (x,y)$ ,

$b = a\tan\theta - \frac{ga^2}{2u^2\cos\theta^2}$

Rearranging the equation, (use $\displaystyle \frac{1}{\cos^2\theta} = \sec^2\theta = 1+\tan^2\theta$ )

$ga^2\tan^2\theta - 2au^2\tan\theta + (ga^2+2y^2b) = 0$

For a real value of $\displaystyle \theta$ to exist,

$D\geq 0$

Hence,

$u^4-2u^2gh-g^2a^2\geq 0$

$\Rightarrow (u^2-gb)^2\geq g^2(b^2+a^2)$

$\Rightarrow \boxed{u \geq \sqrt{gb+g\sqrt{b^2+a^2}}}$

Now, let us encounter our original problem.
alt text
Clearly, the point $\displaystyle (a,b)$ that we ought to consider is $\displaystyle (R,-H)$

Thus,

$u\geq \sqrt{g(-H)+g\sqrt{H^2+R^2}}$

$\Rightarrow \left(\frac{u^2+gH}{g}\right)^2\geq R^2 + H^2$

$\Rightarrow R\leq\sqrt{\left(\frac{u^2+gH}{g}\right)^2-H^2}$

Substituting the values provided in the problem,

$R\leq 2.061m$

$R_{max} = 2.061m$

Therefore,

$159R_{max} = 327.718m \approx \boxed{328m}$

$R=u\cos\theta t$

$-H=u\sin(\theta)t-\frac {1}{2}gt^2$

Now..lets manipulate a bit

$\cos\theta=\frac {R}{ut}$

$\sin\theta=\frac {(\frac {1}{2}gt^2-H)}{ut}$

on substituting values and using the identity

$(\cos\theta)^2+(\sin\theta)^2)=1$

$R^2=4t^2-(4.9t^2-5)^2$

to minimize $R$ ........ minimize $R^2$

on calculating $R_{max}=2.061 m$

Let $\theta$ be the angle with the horizontal at which the particle is thrown. Let $t$ be the time it takes to reach the ground. Let $R$ be the range. Then

$R = u \cos (\theta ) \ t$

$\displaystyle{-H = u \sin (\theta) \ t - \frac {1}{2} g t^{2}}$

Solving for $t$ and substituting (the positive value of t) in the equation for R,

$R = \displaystyle{ \frac {u \cos (\theta) \left [ u \sin(\theta) +\sqrt {(u^{2} \sin^{2}( \theta) + 2gh)} \right ]}{g} }$

For $R_{max}$ find the equation for $\displaystyle{\frac {dR}{d \theta} = 0}$ . Thus we get after some simplification

$\displaystyle{u^{2} (4x^{2} - 4x + 1) ( u^{2}x + 2gh) = x(u^{2} + 2gh)^2 }$

where $x = \sin^{2}( \theta)$ . This is a cubic in x. We get three roots of which two roots are greater than 1. The third root namely $x = 0.0329234$ gives the value for $R_{max} = 2.0609346$ .

I had to obtain the roots for x numerically. Is there a better way to solve this problem in which the solution can be obtained analytically ?

1. Take x-component of the velocity as X and y-component as Y. Therefor, $X^{2} + Y^{2}=2^{2}$ ..... (1)

2. Taking Y as the initial speed, and '-5' as the displacement and '-9.8' as the acceleration, derive an expression for time of flight using the equation $D=ut=\frac{1}{2}at^{2}$ . .....(2)

3. We will get and expression of TIME in terms of Y. Use ...(1) to convert the expression of TIME OF FLIGHT in the terms of X.

4. Multiply T with X, where X is the horizontal velocity. This is an expression of RANGE in terms of X. We have to find the MAXIMUM value of this function of X.

5. Differentiate the expression with respect to X, and equate the differentiation with 0 to get a quadratic equation of X. Find the value of X. Use it to find TIME OF FLIGHT, and then RANGE. :-)

Consider the set of inclined planes starting from the top of the tower to the landing point. My claim is that range on ground will be minimized at the angle that gives minimum range for the corresponding incline.
Max range on incline of angle A= u^2/g(1-sinA)=f(A)

Hence f(A)sinA=H And R=H cotA =2.601 m

Man I really appreciate your effort. But it would be better if you ask greatest integer or integer next to greatest integer ... That would be more specified.. Thanx

Use the vector equation and then used implicit differentiation and set the derivative to 0.