Try it- have 2 click ur minds 2 escape clumsy calculation

As far as I think, the solution will be something like this.

First let consider for any horizontal velocity V_x

The vertical componenet V_y will be minimum when it reduces to 0 at the top most point of tower of height 2H

So the maximum height reached is equal to 2H for this projectile motion.

the second relation can be drawn by the fact that in the time ... V_x covers √3 H ... V_y covers 2H-H = H

I think this will solve the problem.

if u r throwing to a pole h m high d distance away does min velocity is when v_y=0 at pole top?

wont the ans depend upon the dist of the point from the foot of the small tower, from whr the ball is thrown...
the ans i got is dependent on it...

its like.....
if the point of projection is x dist away from small tower..

one point on parabola is (x,H) and the other one is (x+√3H,2H)...

then we apply trajectory equation to both..

we get a quad equation in tanθ(θ=angle of projection)...

in this eqn we just find D>=0 for real u(initial vel.)..

for min u, D=0.

first consider throw frm small tower to long one.

R_max = u²/g(1+sinθ) and this u is min for a given R so

2H = u_min/g(1+sinθ) now u_min is found now apply energy conservation at top os small tower and ground......

nice