Firstly this not a lengthy question can be solved in 2-3 steps.
Two towers of Height H and 2H separated by √3H what is the min velocity required to throw the ball from side of short tower so that it just touches small tower to reach top of second tower.
1As far as I think, the solution will be something like this.
First let consider for any horizontal velocity Vx
The vertical componenet Vy 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 ... Vx covers √3 H ... Vy covers 2H-H = H
I think this will solve the problem.
33if u r throwing to a pole h m high d distance away does min velocity is when vy=0 at pole top?
1the constraint here should be that the ball surpasses the smaller tower n not that max height is 2H....
1wont 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...
1its 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.
33No distance frm the tower and θ would be fixed for min velocity......
Try it i will post the soln tommrow
33first consider throw frm small tower to long one.
Rmax = u2/g(1+sinθ) and this u is min for a given R so
2H = umin/g(1+sinθ) now umin is found now apply energy conservation at top os small tower and ground......
