Solving a Dynamics Problem - Projectile Question

[Auron87]

Bit stuck on a projectile style question, at first glance I thought it would be pretty easy but I'm stuck now!

"A man is shooting another man who has climbed a pole h metres high, L metres away. The bullets are at speed v leaving the gun. Find an equation determining the angle of projection needed (alpha). Show that tan(alpha)=(v^2)gL for the maximum range."

Think I've got the angle part sorted, I have arccos(L/vt) for it anyway! But I'm really struggling getting the next part of the question, the nearest I've got is sin(2alpha) = (gL)/v^2 but can't really get tan into it! Any help would be much appreciated. Thanks.

 

[finchie_88]

Horizontal velocity, displacement and acceleration respectively:
v_x = v_0cos\alpha
s_x = v_0tcos\alpha
a_x = 0

Vertical Velocity, displacement, acceleration respectively:
v_y = v_osin\alpha - gt
s_y = v_0tsin\alpha - \frac{1}{2}gt^2
a_y = -g

\therefore y = \frac{v_0xsin\alpha}{v_0cos\alpha} - \frac{1}{2}g[\frac{x}{v_0cos\alpha}]^2

Cancel down and re-arrange, and you should be able to get the answer. Just is case it is needed, sin2\alpha = 2sin\alpha.cos\alpha and tan\alpha = \frac{sin\alpha}{cos\alpha}
I trust that you can finish.