I long time try, and found this formula:

4*h * n ( n + 1 ) > H

minimal n - is answer!

but I doubt, that this is correct.

It's correct or not?

Re: I long time try, and found this formula:

v0*t*sin(theta)+0.5*g*sin(theta)*t^2 *n>=sqrt(H*H+l*l)

v0=sqrt(2*g*h)

t=2*v0/g

tan(theta)=H/l.

seems to be correct

PS. this problem is too simple..

*Edited by author 12.10.2017 06:25*

*Edited by author 12.10.2017 06:40*

Re: I long time try, and found this formula:

the correct formula is

4*h*n*(n+1)>(H*H+l*l)/H

Re: I long time try, and found this formula:

Maybe provide some explanation

Re: I long time try, and found this formula:

Oh

Your previous comment is the explanation

Re: I long time try, and found this formula:

Let O=(l,H), A=(0,0), B=(l,0). System of coordinates: OX=AO, OY = BO rotate on teta (teta=arctan(H/l)).

Then:

vx(t) = (v+gt)*sin(teta)

vy(t) = (v-gt)*cos(teta)

x(t) = V*sin(teta)+g*sin(teta)*(t^2)/2

y(t) = V*cos(teta)-g*cos(teta)*(t^2)/2

First point: (l, H) => x=0, y=0

Second point: y(t)=0 <=> t=2*v/g

x(2*v/g) = 4*(V^2)*sin(teta)/g = d = 8 * h * sin(teta) (mg(H+h)=mgH+m*(V^2)/2)

x = d, y = 0

Distance between First point and Second point = d

Distance between Second point and Third point = 2*d

...

Distance between i point and i+1 point = i*d

If n = answer => d+2d+3d+...+(n-1)d<=sqrt(H^2 + l^2)

d+2d+3d+...+(n-1)d+nd>sqrt(H^2 + l^2)

*Edited by author 05.08.2018 12:52*